If $C$ is a regular or (Barr-)exact category and $T$ is a Lawvere algebraic theory, then the category $Alg_T$ of $T$-algebras is also regular or exact, respectively. Reference: Exact Categories by Michael Barr, theorem 5.11 (pdf). In particular, if $C$ is exact, then the category of abelian group objects $Ab(C)$ is additive and exact, i.e., is an abelian category.

Also if $C$ is regular/exact, then so is any functor category $C^D$ for $D$ small. So if $C$ is exact, then so is for example $Ab(C^{\Delta^{op}})$, the category of simplicial abelian groups in $C$.

A convenient set of axioms on an abelian category that guarantees enough injectives is given by the notion of Grothendieck category (after the famous Tohoku paper): a cocomplete abelian category with exact filtered colimits and a generator. Suppose that $C$ is a category such that

Then $Ab(C)$ is a cocomplete category with exact filtered colimits. To see this, it helps to know the following facts:

Filtered colimits exist in $Ab(C)$ and are computed just as they are in $C$ (this is true for any category of algebras $Alg_T$).

$Ab(C)$ has arbitrary coproducts (just take a filtered colimit over a system of finite coproducts), hence is a cocomplete abelian category.

About the condition that $Ab(C)$ has a generator: well, if $C$ has a generator $G$ and the underlying functor $U: Ab(C) \to C$ is monadic, then we can construct a free abelian group object $F(G)$ and this will be a generator in $Ab(C)$. Under the relatively strong exactness condition that products distribute over colimits in $C$, one can prove this monadicity (a more general result is given here), but otherwise I'm not sure what general conditions on $C$ guarantee existence of generators in $Ab(C)$. At least this tells us that if $C$ is a Grothendieck topos, then $Ab(C)$ is a Grothendieck category (but maybe you knew that already).

On constructing injective hulls in categories, you may find additional enlightenment in this paper by Barr.