I am self studying Ash & Novinger's Complex Variables. The authors prove the following theorem (see Page 4, Subsection 6.1.6):
Proposition. Let $g_1 , g_2, \ldots$ be a sequence of bounded complex valued functions, each defined on a set $S$. If the series $\sum_{n=1}^{\infty} \lvert g_n \rvert$ converges uniformly on $S$, then the product $\prod_{n=1}^{\infty} (1+g_n)$ converges absolutely and uniformly on $S$. Furthermore, if $f(z)=\prod_{n=1}^{\infty} (1+g_n(z)), z\in S$, then $f(z)=0$ iff $1+g_n(z)=0$ for some $n\in \mathbb N$.
Is the bounded condition redundant? The proof never makes use of it, in fact, uniform convergence of $\sum_{n=1}^{\infty} |g_n|$ implies the boundedness of $g_n$'s after a certain stage. Here's how it can be show:
Let $\sigma_n (x) = \sum_{k=1}^{n} |g_k(z)|$. Since $\sigma_n$ converges uniformly, it is uniformly Cauchy (see here). Thus, there is some $K \in \mathbb N$ such that for every $m,n \ge K$, we have that $|\sigma_{m} (x) - \sigma_{n} (x)| < 1$. The result is now immediate by choosing $m=n+1$ for each $n \ge N$.
Since the first few terms cannot affect convergence, we may in fact get rid of them. So what is the point of boundedness here?
I cannot see why boundedness needs to be required a-priori. As you said, $\sigma_n (z) = \sum_{k=1}^{n} |g_k(z)|$ converges uniformly on $S$, so that $$ |g_n(z)| = \sigma_n (z) - \sigma_{n-1} (z) $$ converges uniformly to zero on $S$, and that implies uniform boundedness of $(g_n)_{n \ge N}$ for some $N$.