Neutral element in $\hom_C(A, B)$

An alternative argument, which might be easier.

If by having abelian group structures on $\operatorname{Hom}_\mathcal{C} (A,B)$ you actually mean that the composition of morphisms is distributive with respect to addition: $$(g+g')\circ f = g\circ f + g'\circ f,\quad g\circ (f+f') = g\circ f + g\circ f'$$ (i.e. that your category is preadditive), then it is enough to note that

• the above identities imply that the neutral elements $0$ with respect to addition of morphisms satisfy $0\circ f = 0$ and $g\circ 0 = 0$ for any $f, g$ (composition of the neutral element with any morphism is the neutral element in the corresponding abelian group);

• the abelian groups $\operatorname{Hom}_\mathcal{C} (A,0) = \{ A\to 0 \}$ and $\operatorname{Hom}_\mathcal{C} (0,B) = \{ 0\to B \}$ are trivial (here $0$ is a zero object, i.e. both initial and terminal), and therefore the composition of arrows $A\to 0$ and $0\to B$ is the neutral element in $\operatorname{Hom}_\mathcal{C} (A,B)$.

So in any preadditive category with a zero object $0$, the neutral element of $\operatorname{Hom}_\mathcal{C} (A,B)$ is the morphism $A\to 0\to B$.