I know this is a really unrigorous answer but this is what I've got. Criticisms are greatly appreciated. I referred to this answer on mathoverflow.
Observe first that $|\mathrm{Hom}\left(\mathbb{Z}^n, G\right)|$ = $|\mathrm{Hom}\left(\mathbb{Z}^{n-1}\times \mathbb{Z}, G\right)|$. Once again, we can uniquely describe each element in that set by describing where they send the bases to; in other words
$$
\Phi_n\left(\gamma_n\right) = \left(\gamma_n\left(1,0,....,0\right), \gamma_n\left(0,1,....,0\right),...,\gamma_n\left(0,0,....,1\right)\right)\quad\text{n times}
$$
But note this is the same as describing where each element sends the first $n-1$ bases to, and then where it sends the $n$th base to. Hence, $\mathrm{im}\left(\Phi_n\right)$ is in bijection with the set of all $\left(\mathrm{im}\left(\Phi_{n-1}\right), g\right)$ where $g\in G$ such that the homomorphism property still holds.
In order for homomorphism to hold, consider an arbitrary $\gamma_{n-1}$ in $\mathrm{Hom}\left(\mathbb{Z}^{n-1}, G\right)$. We have that $\gamma_{n-1}\left(1,0,...,0\right)\ast ...\ast\gamma_{n-1}\left(0,0,...,1\right)$ can commute any way we desire in order to preserve the underlying abelian nature of where the homomorphism is mapping from. Therefore, we require that $\gamma g = g \gamma$. In other words, $g^{-1}\gamma g = \gamma$. Note that this is exactly how we fix objects from $\mathrm{Hom}\left(\mathbb{Z}^{n-1}, G\right)$ by the group action of conjugacy from elements in $G$.
Therefore, consider the group action where $G$ acts on the set $|\mathrm{Hom}\left(\mathbb{Z}^{n-1}, G\right)|$, with conjugacy classes of homomorphisms. Then,
$$
\begin{aligned}
|\mathrm{Hom}\left(\mathbb{Z}^{n-1}\times \mathbb{Z}, G\right)| &= \sum_{g_{n-1}\in G}|\{ \gamma_{n-1}\in \mathrm{Hom}\left(\mathbb{Z}^{n-1}, G\right)\;|\;g\gamma g^{-1} = \gamma \} \\
&= \sum_{g_{n-1}\in G}|\mathrm{Fix}\left(g\right)|
\end{aligned}
$$
Therefore,
$$
|G|\cdot N_{n-1} = \sum_{g_{n-1}\in G}|\mathrm{Fix}\left(g\right)| = |\mathrm{Hom}\left(\mathbb{Z}^{n}, G\right)|
$$
Consider the case where $n=3$, we need to find $N_2$, i.e. the number of conjugacy classes of homomorphisms from $\mathbb{Z}^2$ to $G$. I claim this is $N_1\times N_1$. To loosely argue this, consider that if some $j$ and $k$ are in the same conjugacy class of homomorphisms from $\mathbb{Z}$ to $G$ under operating as $l$, then $\left(j,k\right)$'s are in the same conjugacy class of homomorphisms from $\mathbb{Z}^2$ to $G$ under operating with $l$. Now that we need to count ordered pairs, since in principle $j$ and $k$ could be equal, we need to count it $N_1^2$ times. Hence,
(this part is really sketchy and I don't really know how to formalize it)
$|\mathrm{Hom}\left(\mathbb{Z}^{n}, G\right)| = |G|N^{n-1}$ by induction.
Best Answer
For any $\phi\in IS$, we have $h:=\phi(1)\in H$. We will claim that $$b(h) = i_h=\phi$$ as follows:
$$i_h(n)=h^n=(\phi(1))^n=\phi(n*1)=\phi(n)$$ for all $n\in \mathbb{Z}$. Thus $i_h=\phi$. This means that $\phi=b(h)$ is in the image of $b$. So $b$ is surjective.