= Solution
{c}
Use countably many copies of $A$, and let $c:S=\coprod_{i\geq1}A\to P=\prod_{i\geq1}A$ be the assumed isomorphism. Write $p_i=\pi_ic:S\to A$, let $\delta:A\to P$ be the diagonal with every component $1_A$, and put $q=c^{-1}\delta$. Thus $p_iq=1_A$ for every $i$. Let $s:S\to S$ shift the injections, $s\nu_i=\nu_{i+1}$, and let $\sigma:S\to A$ be the fold map, $\sigma\nu_i=1_A$.
The coordinate relations give $p_1s=0$ and $p_{i+1}s=p_i$, while $p_1\nu_1=1_A$ and $p_{i+1}\nu_1=0$. Therefore $q$ and $\nu_1+sq$ have identical $p_i$-coordinates. Since $c$ is invertible, these coordinates jointly distinguish arrows into $S$, so
$$
q=\nu_1+sq.
$$
The coproduct property gives $\sigma s=\sigma$, because both have value $1_A$ on each injection. Consequently $z=\sigma q$ satisfies
$$
z=\sigma\nu_1+\sigma sq=1_A+z.
$$
This proves the absorbing-endomorphism assertion without assuming an infinite addition operation on hom-sets.
If the category is an <additive category>, each hom-set is an <abelian group>, so cancellation gives $1_A=0$. Any arrow $f:A\to B$ is then $f1_A=f0=0$. Every hom-set has exactly one arrow, and every object is isomorphic to the <zero object>. This is <countable biproducts in an additive category force triviality>:
$$
\boxed{z+1_A=z;\qquad\mathcal C\text{ additive}\ \Longrightarrow\ \mathcal C\simeq\mathbf1.}
$$
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