Direct sum of injective modules is injective
Here's an example of a full exact embedding of the module category of a non-Noetherian ring $S$ into that of a Noetherian ring $R$, preserving all direct sums and direct products. So this gives an example of an abelian subcategory of $R\text{-mod}$ that is not injectively closed, but satisfies many additional "niceness" properties.
Let $k$ be a field, and $R$ the path algebra over $k$ of the quiver with two vertices and three arrows from the first vertex to the second. So an $R$-module is given by the data $(U,V,\alpha,\beta,\gamma)$, where $U$ and $V$ are vector spaces, and $\alpha$, $\beta$ and $\gamma$ are linear maps $U\to V$. Then $R$ is a finite dimensional algebra, and so certainly Noetherian.
Let $S=k\langle x,y\rangle$, the free algebra on two (non-commuting) generators. Then $S$ is not left Noetherian.
The module category of $S$ embeds fully into the module category of $R$ by sending an $S$-module $M$ to the $R$-module given by the data $(M,M,x,y,1_M)$
EDIT Thanks to Jeremy Rickard for several corrective insights in the comments!
In the direction of positive conditions, I'm not sure whether the following conditions are reasonable for your purposes:
Proposition: Let $\mathcal{C}$ be an abelian category which is
locally finitely presentable (i.e. it has a generator of finitely presentable objects and is complete, or equivalently cocomplete) and
"strongly Noetherian" in the sense that any subobject of a finitely-presentable object is finitely-presentable, and similarly for higher presentability degrees.
Then $\mathcal{C}$ is injectively closed.
Proof sketch: One shows that in such a category, an object is injective if and only if it lifts against monomorphisms between finitely-presentable objects. This uses the "strong Noetherian" property. This implies, in conjunction with the Noetherianity property again, that injective objects are closed under filtered colimits; since they are also closed under finite direct sums, they are thus closed under arbitrary direct sums.
Corollary: Let $R$ be a "strongly Noetherian ring" in the sense that $R$-$\mathrm{Mod}$ is "strongly Noetherian", and suppose that $\mathcal{C}$ is a full abelian subcategory of $R$-$\mathrm{Mod}$ which is
closed under kernels and colimits in $R$-$\mathrm{Mod}$, and
generated under colimits by finitely-presentable $R$-modules.
Then $\mathcal{C}$ is injectively closed.
Proof sketch: Check that $\mathcal{C}$ satisfies the hypotheses of the proposition.
Notes:
As Jeremy Rickard points out, for countable $R$, $R$-$\mathrm{Mod}$ is "strongly Noetherian" iff it is Noetherian. So it seems that it is not that much stronger a condition than simply being Noetherian.
In Jeremy Rickard's example, where $\mathcal{C}$ is the subcategory of $R$-$\mathrm{Mod}$ given by the image of $S$-$\mathrm{Mod}$, it is the case that $\mathcal{C}$ is closed under kernels and colimits, and is itself locally finitely presentable. But it doesn't satisfy the hypotheses of the corollary because $\mathcal{C}$ is not generated under colimits by modules which are finitely-presentable as $R$-modules. Another way to say this is that the embedding $S$-$\mathrm{Mod} \to R$-$\mathrm{Mod}$ doesn't preserve finite presentability: the module $S$ itself is finitely-presentable as an $S$-module but not as an $R$-module.