Why is $\mathbb{Z}$ not an inital object of GR or AB?
Claim 1: for every group $G$ there exists a groups morphism from $\mathbb{Z}$ to $G$.
PF: Let $f:\mathbb{Z} \rightarrow G$ be given by: $f(n) = n*1_G$. Clearly $f(1) = 1_G$. Now $f(n+m) = (n+m)*1_G = n*1_G + m*1_G$. Hence $f$ is a groups hom.
Claim 2: $f$ is unique.
PF: This follows from that $\mathbb{Z}$ is generated by $1$ and hence every group hom starting from $\mathbb{Z}$ is completely determined by the image of $1$. And this image has to be $1_G$ by definition of group hom.
Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.
You've mixed up the additive and multiplicative identity of $\mathbb{Z}$. $(\mathbb{Z},+)$ is a group with identity $0$. A group homomorphism from $\mathbb{Z}$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $\mathbb{Z}$ to a given group, in fact, mapping $1 \in \mathbb{Z}$ to any $g \in G$ defines a homomorphism.
The initial and final objects of $\mathbf{Gr}$ and $\mathbf{Ab}$ are the trivial group $\{0\}$.