$\DeclareMathOperator\Ab{Ab}\DeclareMathOperator\Hom{Hom}$If $C$ is any additive category then every object has a unique structure as an abelian group object so $\Ab(C)=C$; but typically $C$ is not abelian. For example, this applies to the category of free abelian groups. One can also think about triangulated categories, which are usually not abelian, although a nice theorem of Freyd gives a canonical embedding in an abelian category. One example that has been studied extensively is the category of spectra in the sense of stable homotopy theory. Similarly, one can consider abelian group objects in the homotopy category of spaces, otherwise known as commutative H-spaces.

The question also says:

In general, I have already trouble to show that $\Hom(A,B)\times \Hom(B,C)\to \Hom(A,C)$ is linear in the left coordinate

Surely this is formal? I have drawn the diagram but sadly I cannot get MathJax to display it.

**Update:**

Just to be clear about notation, I'll write $\mathcal{C}(X,Y)$ for morphism sets in $\mathcal{C}$, and $\Hom(A,B)$ for morphism sets in $Ab(\mathcal{C})$. An object $A\in \Ab(\mathcal{C})$ has a natural abelian group structure on $\mathcal{C}(T,A)$ for all $T\in\mathcal{C}$. Naturality means that $q\circ p+r\circ p=(q+r)\circ p$ for all $p:S\to T$ and $q,r:T\to A$. Now let $B$ be another object of $\Ab(\mathcal{C})$. A morphism in $\Ab(\mathcal{C})$ from $A$ to $B$ is just a morphism $f:A\to B$ in $\mathcal{C}$ with the property that $f\circ(p+q)=f\circ p+f\circ q$ for all $T$ and all $p,q\in\mathcal{C}(T,A)$. Now suppose we have such morphisms $f,g:A\to B$ and $h,k\:B\to C$. We then have

\begin{multline*} (f+g)\circ(p+q) = f\circ(p+q) + g\circ(p+q) =
f\circ p + g\circ p + f\circ q + g\circ q ={}
\\
(f+g)\circ p + (f+g)\circ q
\end{multline*}

(using the naturality of addition, the homomorphism property of $f$ and $g$, and then naturality again). This shows that $f+g$ is again a homomorphism. A similar argument shows that $h\circ f$, $h\circ g$ and $h\circ(f+g)$ are homomorphisms. We have $h\circ(f+g)=h\circ f+h\circ g$ by the homomorphism property of $h$. We also have $(h+k)\circ f=h\circ f+k\circ f$ by the naturality of addition.