The theorems you describe only consider products of the form $\prod(1+a_n)$ with $a_n \geqslant 0$ and $\prod(1-b_n)$ with $0\leqslant b_n < 1$.
In general, we have infinite products where $a_n$ could be any real or complex number. The strongest condition that guarantees convergence of $\prod(1+a_n)$ is, of course, absolute convergence. This reverts back to your first theorem.
The product $\prod(1+a_n)$ where $a_n \in \mathbb{R} \text{ or }
\mathbb{C}$ is said to be absolutely convergent if $\prod (1+|a_n|)$
is convergent. The product converges absolutely if and only if
$\sum|a_n|$ is convergent. Also, convergence of $\prod (1+|a_n|) $
implies convergence of $\prod (1 +a_n)$.
In the absence of absolute convergence, the product may still converge conditionally but, as far as I know, there are no all-encompassing necessary and sufficient conditions.
There are some ad hoc theorems for conditional convergence of products where the sign of $a_n$ is unrestricted. For example:
If the series $\sum a_n^2$ is convergent, then the product
$\prod(1+a_n)$ and the series $\sum a_n$ either both converge or both
diverge.
To prove this note that if $\sum a_n^2$ converges then $|a_n| \to 0$ and for sufficiently large $n$ we have $|a_n| < 1/2$ and
$$|\log(1+a_n) - a_n| = \left|\sum_{k=2}^\infty(-1)^{k}\frac{a_n^k}{k} \right| \leqslant \frac{a_n^2}{2}\sum_{k=0}^\infty|a_n|^k = \frac{a_n^2}{2}\frac{1}{1-|a_n|}< a_n^2$$
Thus the series $\sum [\log(1+a_n) - a_n]$ is absolutely convergent by the comparison test, and we have existence of the limit,
$$\lim_{N\to \infty} \log \left(\prod_{n=1}^N (1+a_n)\right) - \sum_{n=1}^Na_n =\lim_{N\to \infty} \prod_{n=1}^N [\log(1+a_n)- a_n ] $$
proving that the product and series must converge or diverge together.
Best Answer
I shall try to give examples where $\sum|a_n|^2$ is divergent and all possible combinations of convergence/divergence for $\prod(1+a_n)$ and $\sum a_n$.
Let $a_{2n}=\frac1{\sqrt n}$ and $a_{2n+1}=\frac1{1+a_{2n}}-1=-\frac{1}{1+\sqrt n}$. Then $(1+a_{2n})(1+a_{2n+1})= 1$, hence the product converges. But $a_{2n}+a_{2n+1}=\frac1{n+\sqrt n}>\frac1{2n}$, hence $\sum a_n$ diverges.
Let $a_{2n}=\frac1{\sqrt n}$ and $a_{2n+1}=-\frac1{\sqrt n}$. Then $a_{2n}+a_{2n+1}=0$, hence $\sum a_n$ converges. But $(1+a_{2n})(1+a_{2n+1})=1-\frac1n$; the $\log$ of this is $\sim -\frac1n$, hence $\sum \log(1+a_n)$ and also $\prod(1+a_n)$ diverges.
Let $a_n=\frac1{\sqrt n}$. Then $\prod(1+a_n)$ and $\sum a_n$ diverge.
It almost looks as if it is not possible to have both $\prod(1+a_n)$ and $\sum a_n$ convergent if $\sum |a_n|^2$ diverges because $\ln(1+a_n) = a_n-\frac12a_n^2\pm\ldots$, but here we go: If $n=4k+r$ with $r\in\{0,1,2,3\}$, let $a_n = \frac{i^r}{\sqrt k}$. Then the product of four such consecutive terms is $(1+\frac1{\sqrt k})(1+\frac i{\sqrt k})(1-\frac1{\sqrt k})(1-\frac i{\sqrt k})=1-\frac1{k^2}$, hence the log of these is $\sim -\frac1{k^2}$ and the product converges. The sum also converges (to $0$).