Let $C$ be a category with all finite (co)limits and a $0$ object. Let $\mathbf{Ab}(C)$ be the category of abelian group objects in $C$, i.e. the category of representable functors $C^{op} \to Set$ that factor through the category of abelian groups. Does $\mathbf{Ab}(C)$ have all finite (co)limits?
Equivalently, does a finite (co)limit of objects in $C$ each with an abelian group structure have a group structure? Obviously, one must assume that all the maps are group homomorphisms.
On
Another way to approach the case of finite limits existing is as follows.
Abelian group objects in $C$ are models of a finite limit sketch.
If you're not familiar with the term, the least technical description is to let $\mathbf{T}$ be the opposite category of the category of finitely presented groups. Then an abelian group object in $C$ is the same thing as a functor $M : \mathbf{T} \to C$ that is left exact — i.e. preserves finite limits.
(The underlying object is $M(\mathbb{Z})$. Addition is the map $M(\mathbb{Z}^2) \cong M(\mathbb{Z})^2 \to M(\mathbb{Z})$ induced by the diagonal $\mathbb{Z} \to \mathbb{Z}^2$. And similarly for the other operations)
Since $C$ has finite limits, finite limits in the functor category $C^{\mathbf{T}}$ are computed pointwise. And we can see that a limit of left exact functors is left exact by the following argument.
Let $F = \lim_i F_i$ be a finite limit of functors and $t = \lim_j t_j$ be a finite limit of objects of $\mathbf{T}$. Assuming the $F_i$ preserve finite limits, interchange of limits proves that $F$ does as well:
$$\begin{align} F(t) &= (\lim_i F_i)(t) = \lim_i F_i(t) \\&= \lim_i \left( F_i(\lim_j t_j) \right) = \lim_i \left( \lim_j F_i(t_j) \right) \\&= \lim_j \left( \lim_i F_i(t_j) \right) = \lim_j \left( (\lim_i F_i)(t_j) \right) \\&= \lim_j F(t_j) \end{align}$$
Note that if $C$ has all small limits, then the same argument applies to show $\mathbf{Ab}(C)$ has all small limits.
Also of note is that if $C$ has filtered colimits that commute with finite limits, then $\mathbf{Ab}(C)$ also has filtered colimits that commute with finite limits.
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.)