Prove a result by assuming it's true and showing no contradiction

Your proof reminds me of the many 'proofs' I have seen for induction, where instead of showing that:

$$\varphi(0)=[some \ formula]$$

or

$$\varphi(n+1)=[some \ formula]$$

it is instead assumed, and then through various manipulations one ends up with $0=0$, which is then followed by the remark: "See, it checks out!"

The big danger with this is that if you cannot reverse the process, then it is not a proof at all .... Indeed, showing that $0=0$ shows nothing at all ... of course $0=0$ is true!

Example: Suppose I want to prove that for all $n$:

$\Sigma_{i=0}^n i = 42$

(which is of course false!)

OK, so I say: Well, for the base I need to show that $\Sigma_{i=0}^0 i = 42$

Well, let's see:

$\Sigma_{i=0}^0 i = 42$

$0*\Sigma_{i=0}^0 i = 0*42$

$0=0$

"Aha!! QED!"

...except it's not QED at all ... Sure, you can always derive $0=0$. And, for that matter, you can always derive $a=a$. None of that shows anything at all.

So yes, you are right that you really want to reverse this process to count as a proof of $i * a = a$

OK, but if so, it seems like you could just start with $a * a = a* a$. That is, you'd get:

$$ a*a = a*a $$ (This is OK to start with, since it's always true) $$ (a*i)*a=a*a $$ (since we're given that $a*i=a$) $$ a*(i*a)=a*a $$ (because of Association) $$ i*a=a $$ (Hmmmm ... why does that follow from $ a*(i*a)=a*a $? That is not at all clear!)

One more thing: The tile of your question raises red flags as well: just because some assumption does not lead to a contradiction does not mean that it is true ... or implied by the givens.

Example: I give you that $a=1$. Now, is $b=2$? If you assume $b=2$, no contradiction will follow ... but it is clearly not implied by $a=1$

Also realize that you being unable to derive a contradiction does not mean there is no contradiction.... maybe there is a contradiction, but you just didn't see how to derive it.

And finally, just because you are able to infer $a=a$ does not mean that there is no contradiction. Again, you can always infer $a=a$. .. including from a contradiction ... indeed, anything follows from a contradiction.

(First some vocabulary. If $a*i = a; \forall a \in S$ then $i$ is a right identity and if $u*a = a;\forall a \in S$ then $u$ is a left identity. If $aa^{-1} = i$ where $i$ is a right identity then $a^{-1}$ is a right inverse and if $a^!a = u$ where $u$ is a left inverse.)

First: You are correct that a proof from conclusion to non-contradiction is not valid unless every step is reversible.

In this proof the irreversible step is the very first. $a = ia \implies a*a = a*ia$ but $a*a = a*ia \not \implies a = ia$.

(But the statement $a*a = (a*i)*a = a*(i*a)$ is most certainly correct.)

Second: The axioms of a group INCLUDE that $a*i = i*a = a;\forall a\in G$. so there is nothing to prove.

There are (among others) two axioms of a group. 1) that there is an identity element that is both a right and left identity. And 2) that for every element $a\in G$ there exists an element that if both a left and right inverse of $a$.

If we were to replace axiom 1) with simply that there is a right identity element that may not be a left identity but keep axiom 2) in tact then:

The step in the proof above is reversible. sort of. $i = a^{-1}a = aa^{-1}$. so $a*a = a*(ia)\implies a^{-1}*a*a = a^{-1}*a*(ia)\implies a*a^{-1}*a = i*(i*a)\implies a*i = (i*i)*a \implies a = i*a$. And we have proven the stronger axiom 1)

Likewise if we were to replace axiom 2) with every element has a right identity but not necessarily a left inverse, but kept axiom 1) in tact we could prove the stronger 2) via

$i = hh^{-1}$ so $(h^{-1})^{-1}=i(h^{-1})^{-1} = hh^{-1}(h^{-1})^{-1}=h*i = h$. So $h^{-1} h = h^{-1}(h^{-1})^{-1} = i= hh^{-1}$

But if we replace BOTH 1) and 2) with a weaker right identities exist and every element has a right inverse but say nothing of left inverses or left inverses, we can not prove a left inverse exist.

For a counter example look up a semigroup with a right identity and right inverse but not left indentities.

(Worth noting that a structure can have a right identity but not a left identity or a left identity but not right identity, but if a structure has both a left and right identity they must be same element. [If a structure has a left identity $u$ and right identity $i$ then $u*i = u$ because $i$ is a right identity and $u*i = i$ because $u$ is a left identity.])