The functions $f_{n}(z)$ are analytic. My work so far is as follows. Since $1+|f_{n}(z)|\le e^{|f_{n}(z)|}$ (it's true that $|f_{n}(z)|\ge 0$), $$\begin{align*}\displaystyle\prod_{n}(1+|f_{n}(z)|)&\le \displaystyle\prod_{n}e^{|f_{n}(z)|}\\ \displaystyle\prod_{n}(1+|f_{n}(z)|)&\le e^{\sum_{n}|f_{n}(z)|}.\end{align*}$$
I've got that $\sum_{n}|f_{n}(z)|$ converges if and only if $\prod_{n}(1+|f_{n}(z)|)$ converges. Now I don't know how to show that if $\sum_{n}|f_{n}(z)|$ converges on each compact subset of $\mathbb C$, then $\prod_{n}(1+f_{n}(z))$ converges on each compact subset of $\mathbb C$.
choosing a suitable branch of log function (like principal branch) make it invertible with continuous(even differentiable) invers function. You can do it since for large enough $m$ the sequence $a_{N} = \prod_{n=m}^{N} \ (1+f_n) $ has positive real part on the compact set, let say $B$ (to argue this you need to use the assumption that $\sum_{n}|f_{n}(z)| < \infty $).
Now fix $z$ by taking $\log$ we get, $\quad \log a_N = \sum_{n=m}^{N} \log (1+f_n (z)) .$ Note that the above series is absolutely convergent because
$$ \sum_{n=m}^{N} \big |\log (1+f_n (z))\big| \leq \sum_{n}|f_{n}(z)| < \infty $$
Thus the sequence $\log a_N$ is convergent to a number like $\ell \neq 0$. This implies $a_N$ itself is convergent to a complex number, due to invertibility of this branch of $\log$ with continuous inverse.