In the real setting, the product $\prod (1 + b_n)$ and the series $\sum b_n$ converge or diverge together for the separate cases where $b_n \geqslant 0$ and $-1 < b_n < 0$ for all $n$.
Even in the real setting, we can find a sequence $(b_n)$ such that the product and series are not simultaneously convergent if we allow $b_n$ to change sign.
For example, let
$$b_{2n-1}= -\frac{1}{\sqrt{n+1}}, \\ b_{2n}= \frac{1}{\sqrt{n+1}}+\frac{1}{n+1}+ \frac{1}{\sqrt{n+1}(n+1)}. $$
We have
$$(1 + b_{2n-1})(1 + b_{2n})= 1 - \frac{1}{(n+1)^2},$$
and
$$P_{2m}= \prod_{n=1}^{m}\left(1+ \frac{1}{(n+1)^2} \right), \\ P_{2m+1}= P_{2m}\left(1 - \frac{1}{\sqrt{m+2}}\right).$$
The partial product $P_{2m}$ converges since the series $\sum(n+1)^{-2}$ converges.
Furthermore
$$\lim_{m \to \infty} P_{2m+1} = \lim_{m \to \infty} P_{2m}\lim_{m \to \infty}\left(1 - \frac{1}{\sqrt{m+2}}\right) = \lim_{m \to \infty} P_{2m}.$$
This shows that the product $\prod (1 + b_n)$ is convergent as partial products with both odd and even number of factors converge to the same limit.
However, the series $\sum b_n$ diverges analogously to the harmonic series.