You may know that convergence of products is not defined just in terms of "limit of partial products exists", for example $\prod \frac 1n$ is by definition not convergent, even though the partial products converge to $0$. Instead the definition is such that there may be finitely many factors $=0$ and for the remaining factors the sum of the logs is convergent. If we get rid of the logarithm, this gives us the direct definition:
$\prod x_n$ is said to be convergent if there exists $N$ such that the sequence of partial products $\prod_{n=N}^Mx_n$ converges to a nonzero(!) limit as $M\to\infty$.
This is also why absolut convergence of $\prod (1+a_n)$ infinite products is defined as absolute convergence of $\sum\ln(a+a_n)$ (ignoring the possibly finitely many factors $=0$ here). (After all, the signs of the factors themselves cannot play a role: they must converge to $+1$, hence there can only be finitely many negative factors.) As a consequence, we have the same properties as with sums, i.e., absolute convergence implies that the order of factors does not matter.
Now assume $\sum \ln(1+a_n)$ is convergent. Then specifically $a_n\to 0$. As $\ln(1+x)=x+O(x^2)$ we see that the behaviour of $\sum \ln(1+a_n)$ is essentially the same as that of $\sum a_n$. Therefore the convergence/absolute convergence of $\prod(1+a_n)$ can be related to convergence/absolute convergence of $\sum a_n$.