Can we use the internal logic of a category to do diagram chases "as in $\mathbf{Ab}$" ?
This MathOverflow question and answer and the referenced slides and the comments both by Ingo Blechschmidt seem directly aimed at your question. See also the nLab page on diagram chasing whose section on diagram chasing in abelian categories addresses the approach you are considering. What you are suggesting is the fourth approach. For your purposes, you may want to look at the other approaches mentioned as well, such as the two element in an abelian category approaches.
As far as I can tell from the above, the "obvious" approach works. Basically, an abelian category is a regular category, so you can definitely interpret regular logic. You can then assert the existence of an abelian group structure for every sort/type, i.e. $\cfrac{\Gamma\vdash t_1:B\qquad \Gamma\vdash t_2:B}{\Gamma\vdash t_1+t_2 : B}$ and so forth. The mentioned "axiom of unique choice", true in any lex category, states that morphisms are equivalent to total functional relations in the internal logic. With the caveat of requiring constructive logic, you can use ordinary element-wise reasoning in this language and the results will be true in any abelian category.
Starting with a regular logic and then asserting every sort has an abelian group structure, should essentially give you the internal language of a regular additive category. However, an abelian category is precisely an exact additive category. Regular categories and exact categories are intimately related. A regular category says all congruences of the form $\Gamma,x:A,y:A\vdash f(x)=f(y)$ for any morphism $f$ have quotients, i.e. coequalizers of the two components of $\{\langle x,y\rangle\in A\times A\mid f(x)=f(y)\}\rightarrowtail A\times A$. This is exactly the image of $f$. An exact category states that every congruence is of the above form for some $f$. (In particular, for the quotient map $q:A\to A/{\sim}$.)
I'm just making this up and patching things together, so take this all with a large grain of salt, but combining the presentation of a regular type theory with effective quotients in Maria Maietti's Modular correspondence between dependent type theories and categories including pretopoi and topoi should give a characterization of exactness. Then adding the additive group structure should give a type theory compatible with abelian categories. I adapt Maietti's presentation notationally. I'll use $\Gamma\vdash T\ \mathsf{prop}$ to mean $\Gamma\vdash T\ \mathsf{type}$ and $\Gamma,x:T,y:T\vdash x=y:T$, i.e. that $T$ is a mono type. Terms of mono types correspond to subobjects. As another shorthand, I'll write $\Gamma\vdash T$ when $T$ is a monotype to mean $\Gamma\vdash c:T$ where $c$ is any term of type $T$ in scope (which are necessarily all equal). I'm omitting the usual structural rules and the rules for equality.
Terminal object: $$\cfrac{}{\Gamma\vdash\top\ \mathsf{type}}\top F\qquad\cfrac{}{\Gamma\vdash\star:\top}\top I \qquad\cfrac{\Gamma\vdash t:\top}{\Gamma\vdash t = \star:\top}\top\beta$$
Dependent sum (and thus finite products): $$\cfrac{\Gamma,x:B\vdash C(x)\ \mathsf{type}}{\Gamma\vdash\Sigma x:B.C(x)\ \mathsf{type}}\Sigma F\qquad\cfrac{\Gamma\vdash b:B\qquad\Gamma\vdash c:C(b)\qquad\Sigma x:B.C(x)\ \mathsf{type}}{\Gamma\vdash\langle b,c\rangle:\Sigma x:B.C(x)}\Sigma I$$
$$\cfrac{\Gamma,z:\Sigma x:B.C(x)\vdash M(z)\ \mathsf{type}\qquad \Gamma\vdash d:\Sigma x:B.C(x)\qquad\Gamma,x:B,y:C(x)\vdash m(x,y):M(\langle x,y\rangle)}{\Gamma\vdash\mathsf{elim}_\Sigma(d,x,y.m(x,y)):M(d)}\Sigma E$$
$$\cfrac{\Gamma,z:\Sigma x:B.C(x)\vdash M(z)\ \mathsf{type}\qquad \Gamma\vdash b:B\qquad \Gamma\vdash c:C(b)\qquad\Gamma,x:B,y:C(x)\vdash m(x,y):M(\langle x,y\rangle)}{\Gamma\vdash\mathsf{elim}_\Sigma(\langle b,c\rangle, x,y.m(x,y))=m(b,c):M(\langle b,c\rangle)}\Sigma\beta$$
Extensional equality type (and thus pullbacks given the dependent sum): $$\cfrac{\Gamma\vdash C\ \mathsf{type}\qquad\Gamma\vdash c:C\qquad\Gamma\vdash d:C}{\Gamma\vdash\mathsf{Eq}(C,c,d)\ \mathsf{type}}\mathsf{Eq}F\qquad\cfrac{\Gamma\vdash c:C}{\Gamma\vdash\mathsf{refl}_C(c):\mathsf{Eq}(C,c,c)}\mathsf{Eq}I$$
$$\cfrac{\Gamma\vdash p:\mathsf{Eq}(C,c,d)}{\Gamma\vdash c=d:C}\mathsf{Eq}E\qquad\cfrac{\Gamma\vdash p:\mathsf{Eq}(C,c,d)}{\Gamma\vdash p=\mathsf{refl}_C(c):\mathsf{Eq}(C,c,d)}\mathsf{Eq}\beta$$
Existential type (which can actually be defined as $(\Sigma x:B.C(x))/\top$): $$\cfrac{\Gamma,x:B\vdash C(x)\ \mathsf{prop}}{\Gamma\vdash\exists x:B.C(x)\ \mathsf{type}}\exists F\qquad\cfrac{\Gamma\vdash b:B\qquad\Gamma\vdash c:C(b)\qquad\Gamma\vdash\exists x:B.C(x)\ \mathsf{type}}{\Gamma\vdash(b,c):\exists x:B.C(x)}\exists I$$
$$\cfrac{\Gamma\vdash\exists x:B.C(x)\ \mathsf{type}\qquad\Gamma\vdash b:B\qquad\Gamma\vdash c:C(b)\qquad\Gamma\vdash d:B\qquad\Gamma\vdash t:C(d)}{\Gamma\vdash(b,c)=(d,t):\exists x:B.C(x)}\exists{=}$$
$$\cfrac{\Gamma,z:\exists x:B.C(x)\vdash M(z)\ \mathsf{prop}\qquad\Gamma\vdash d:\exists x:B.C(x)\qquad\Gamma,x:B,y:C(x)\vdash m(x,y):M((x,y))}{\Gamma\vdash\mathsf{elim}_\exists(d,x,y.m(x,y)):M(d)}\exists E$$
Effective extensional quotient types: $$\cfrac{\Gamma,x:A,y:A\vdash R(x,y)\ \mathsf{prop}\qquad\Gamma,x:A\vdash R(x,x)\qquad\Gamma,x:A,y:A,p:R(x,y)\vdash R(y,x)\qquad\Gamma,x:A,y:A,z:A,p:R(x,y),q:R(y,z)\vdash R(x,z)}{\Gamma\vdash A/R\ \mathsf{type}}QF$$
$$\cfrac{\Gamma\vdash a:A\qquad\Gamma\vdash A/R\ \mathsf{type}}{\Gamma\vdash[a]:A/R}QI\qquad\cfrac{\Gamma\vdash a:A\qquad\Gamma\vdash b:A\qquad\Gamma\vdash A/R\ \mathsf{type}}{\Gamma\vdash [a]=[b]:A/R}Q{=}$$
$$\cfrac{\Gamma,z:A/R\vdash L(z)\ \mathsf{type}\qquad\Gamma\vdash p:A/R\qquad\Gamma,x:A\vdash l(x):L([x])\qquad\Gamma,x:A,y:A,d:R(x,y)\vdash l(x)=l(y):L([x])}{\Gamma\vdash\mathsf{elim}_Q(p, x.l(x)):L(p)}QE$$
$$\cfrac{\Gamma,z:A/R\vdash L(z)\ \mathsf{type}\qquad\Gamma\vdash a:A\qquad\Gamma,x:A\vdash l(x):L([x])\qquad\Gamma,x:A,y:A,d:R(x,y)\vdash l(x)=l(y):L([x])}{\Gamma\vdash\mathsf{elim}_Q([a], x.l(x))=l(a):L([a])}Q\beta$$
$$\cfrac{\Gamma\vdash a:A\qquad\Gamma\vdash b:A\qquad\Gamma\vdash [a]=[b]:A/R}{\Gamma\vdash\mathsf{eff}(a,b):R(a,b)}\mathsf{Eff}$$
Additive structure: $$\cfrac{\vdash A\ \mathsf{type}}{\Gamma\vdash 0_A:A}0I\qquad\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A\qquad\Gamma\vdash b:A}{\Gamma\vdash a+b:A}{+}I\qquad\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A}{\Gamma\vdash -a:A}{-}I$$
$$\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A\qquad\Gamma\vdash b:A}{\Gamma\vdash a+b=b+a:A}{+}\sigma\qquad\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A}{\Gamma\vdash 0+a=a:A}{+}\lambda$$
$$\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A\qquad\Gamma\vdash b:A\qquad\Gamma\vdash c:A}{\Gamma\vdash (a+b)+c=a+(b+c):A}{+}\alpha\qquad\cfrac{\vdash A\ \mathsf{type}\qquad\Gamma\vdash a:A}{\Gamma\vdash a+(-a)=0_A:A}{+}i$$
$$\cfrac{\vdash A\ \mathsf{type}\qquad\vdash B\ \mathsf{type}\qquad x:A\vdash t:B}{\Gamma\vdash t[0_A/x]=0_B:B}0c$$
$$\cfrac{\vdash A \ \mathsf{type}\qquad\vdash B\ \mathsf{type}\qquad\Gamma\vdash a:A\qquad\Gamma\vdash b:A\qquad x:A\vdash t:B}{\Gamma\vdash t[a+b/x]=t[a/x]+t[b/x]:B}{+}c$$
It is not the case that $\top$ and $A\times B(=\Sigma x:A.B)$ correspond to a void/empty/false/$\bot$ type and coproduct type respectively, because initial objects and coproducts in abelian categories are not stable, i.e. preserved by pullback functors. What we can show is that they do behave like void types and coproduct types if we only consider closed types, i.e. the types that correspond to objects of the abelian category rather than objects in a fiber category. All the additive structure is restricted to terms of closed types.
Let $\vdash A\ \mathsf{type}$, $\vdash B\ \mathsf{type}$, and $\vdash C\ \mathsf{type}$, i.e $A$, $B$, and $C$ are closed types, from this point on. Let's show that $\top$ is initial. Clearly, $x:\top\vdash 0_A:A$ shows that there exists an arrow from $\top$ to any (closed) $A$. To show uniqueness, consider a term $x:\top\vdash t(x):A$. Since $x=\star:\top$ via $\top\beta$, this term is the same as $x:\top\vdash t(\star):A$ and so we have $t(x)=t(\star)=t[t'/x]$ for any $t':\top$. In particular, for $t'=0_\top$ we get $t(x)=t[0_\top/x]=0_A$ via $0c$. (Really, we could replace the rule $0c$ with the rule $\frac{\vdash t:A}{\Gamma\vdash t=0_A:A}$ which is to say the only constant term is zero which makes sense as the only constant abelian group homomorphism is the zero morphism.) Next, write $A\oplus B$ for $A\times B$. Define $\mathsf{inl}(a)$ as $\langle a,0_B\rangle$ and $\mathsf{inr}(b)$ as $\langle 0_A,b\rangle$. Define $\mathsf{elim}_\oplus(t,x.f(x),y.g(y))$ as $\mathsf{elim}_\Sigma(t, x,y.f(x)+g(y))$ where $\Gamma\vdash t:A\oplus B$, $\Gamma,x:A\vdash f(x):C$, and $\Gamma,y:B\vdash g(y):C$. It's easy to show that this satisfies the universal property of the coproduct.
The kernel of $\Gamma,x:A\vdash f(x):B$ is $\Sigma x:A.\mathsf{Eq}(B,f(x),0_B)$. The image is $A/R$ where $R(x,y)$ is defined as $\mathsf{Eq}(B,f(x),f(y))$. The cokernel is $B/R$ where $R(x,y)$ is defined as $\exists a:A.\mathsf{Eq}(B,f(a),x-y)$. The coimage is $A/R$ where $R(x,y)$ is defined as $\exists a:A.\mathsf{Eq}(B,f(a),0_B)\times\mathsf{Eq}(A,a,x-y)$. Showing that the image is isomorphic to the kernel of the cokernel using this language is a good exercise. Similarly for the coimage and the cokernel of the kernel. From here it is fairly obvious that the coimage is isomorphic to the image. Either direction is just $\mathsf{elim}_Q(z,x.[x])$ whose well-formedness in both directions just requires $(\exists a:A.f(a)=0\land a = x-y)\iff (f(x)=f(y))$ which clearly holds.