That is not what Euler's product expansion of the sine looks like. It is very much supposed to be in the form $\frac{\sin z}{z} = \prod_{n \ge 1} \left( 1 - \frac{z^2}{\pi^2 n^2} \right).$
The product you've written down does not converge for $A = \{ n \pi \}$ unless $z \in A$. Indeed, its factors don't go to $1$, which is a necessary condition exactly analogous to the condition for infinite series that the terms need to go to $0$. In fact one can switch between infinite sums and infinite products using the logarithm, which can be used to prove the following.
(First I need to mention that the theorem below requires the convention wherein a product which tends to $0$ is said to diverge. This is because the logarithm of such a product diverges to $-\infty$.)
Theorem: Let $a_n \in \mathbb{C}$ be a sequence such that $\sum |a_n|^2$ converges. Then $\prod (1 + a_n)$ converges if and only if $\sum a_n$ converges.
Sketch. Use the fact that $\log (1 + a_n) = a_n + O(|a_n|^2)$.
So we can make sense of the "infinite polynomial"
$\prod_{\alpha \in A} \left( 1 - \frac{z}{\alpha} \right)$
for countable $A$ such that $\sum_{\alpha \in A} \frac{1}{|\alpha|^2}$ and $\sum_{\alpha \in A} \frac{1}{\alpha}$ both converge. See also the Weierstrass factorization theorem.
Note that by the identity theorem, the zeroes of a holomorphic function are isolated, so if you want your product to be holomorphic with $A$ as its zero set, $A$ needs to be discrete.
Infinite sums and products do not behave well for uncountably many terms, the basic reason being the following.
Theorem: Let $S$ be an uncountable set of positive real numbers. Then for any positive real $r$, there is a finite subset of $S$ whose sum is greater than $r$.
(In other words, no sum with uncountably many terms can converge absolutely.)
Proof. The sets $S_{\epsilon} = \{ s : s \in S, s > \epsilon \}$ for $\epsilon$ a positive rational are a countable collection of sets whose union is $S$. Since a countable union of countable sets is countable, it follows that there exists $\epsilon$ such that $S_{\epsilon}$ is uncountable. Then the result is clear.