(Co)completeness of the category of abelian group objects in a (co)complete category

For limits, yes. The functor category $Ab^{C^{op}}$ has limits (evaluated pointwise). Moreover, in $Set^{C^{op}}$, a finite limit of representable functors is representable, since an object representing the limit is the same thing as a limit of the representing objects, and $C$ has all finite limits. This remains true in $Ab^{C^{op}}$ since limits are evaluated pointwise in both categories and the forgetful functor $Ab\to Set$ creates limits. Thus $\mathbf{Ab}(C)$ has finite limits, and more generally the forgetful functor $\mathbf{Ab}(C)\to C$ creates limits.

Or if you prefer, in more concrete terms: if you take a limit of a diagram of abelian group objects, the set of maps from any other object $X$ of $C$ into the limit will be the limit of the sets of maps from $X$ to each object in the diagram (this is basically the definition of a limit). Since the limit (in $Set$) of a diagram of abelian groups and homomorphisms has a natural abelian group structure, this means the limit object is an abelian group object.

For colimits the story is not so nice. We see this already in the case $C=Set$, where the forgetful functor $Ab\to Set$ fails to preserve colimits (although colimits do exist in $Ab$). For an example where finite colimits exist in $C$ but not in $\mathbf{Ab}(C)$, let $C$ be the poset $\mathbb{N}$. Then $\mathbf{Ab}(C)$ is the empty category, since if $X$ is an abelian group object then $\operatorname{Hom}_C(Y,X)$ is an abelian group and in particular is nonempty for all $Y$, but there is no object of $C$ that has a map from every other object. In particular, $\mathbf{Ab}(C)$ does not have an initial object.

(However, if $C$ has finite limits, then $\mathbf{Ab}(C)$ at least has finite coproducts, since finite products and finite coproducts are the same in $\mathbf{Ab}(C)$ and more generally in any category enriched in abelian groups. I don't know of an example where $C$ has both finite colimits and finite limits but $\mathbf{Ab}(C)$ fails to have coequalizers, but I think that should be possible.)