$\mathbb{Z}^n$ is typically defined as the direct sum of $n$ copies of $\mathbb{Z}$. Also known as the free abelian group.
On the other hand, the free product of $n$ copies of $\mathbb{Z}$, which you denote by $A_n$, is isomorphic to the free group $F_n$ (note: without "abelian") on $n$ generators.
These groups are isomorphic only for $n=1$. When $n>1$ then $F_n$ is not abelian while $\mathbb{Z}^n$ is. Formally presentations are as follows:
$$F_n=\big\langle \{g_1,\ldots, g_n\}\ \big|\ \emptyset \big\rangle$$
$$\mathbb{Z}^n=\big\langle \{g_1,\ldots, g_n\}\ \big|\ \{g_ig_kg_i^{-1}g_k^{-1}\}_{i,k=1}^n\big\rangle$$
Or a bit less formally:
$$F_n=\big\langle g_1,\ldots, g_n\big\rangle$$
$$\mathbb{Z}^n=\big\langle g_1,\ldots, g_n\ \big|\ g_ig_k=g_kg_i\text{ for any }i,k\big\rangle$$
So as you can see they don't have the same presentation: $\mathbb{Z}^n$ has additional "commutativity" relations. Besides they can't have the same presentation, because the same presentation implies isomorphism. Unless someone approaches the topic in a non-standard way (e.g. ignoring commutativity relations), but then he/she should be explicit about it. It is very unlikely for a typical mathematician to think that $\mathbb{Z}^n$ and $F_n$ have the same presentation.
$A*B$ is never co-hopfian, as it contains as a proper (isomorphic) subgroup $A*(ba)B{(ba)}^{-1}$, where $a, b$ are two arbitrary non trivial elements in $A$ and $B$ respectively.
Best Answer
Yes.
It's a general pattern based on the free-forgetful adjunction: the free functor preserves colimits, in particular coproducts, so since $S=\coprod_{s\in S}1$ where $1$ is a set with one element, then $$F(S)=\coprod_{s\in S}F(1)\,,$$ and the coproduct in the category of groups is free product and $F(1)=\Bbb Z$.