On Ahlfors on p. 191 he is talking about the relation between $\prod_{n=1}^\infty (1+a_n)$ and $\sum_{n=1}^\infty \log(1+a_n)$. He says:
Since the $a_n$ are complex, we must agree on a definite branch of the logarithms, and we decide to choose the principal branch in each term.
He makes no mention of the possibility that one of the terms might lie on the negative real axis. What then?
Since a necessary condition for the product to converge is that the factors converge to $1$, when testing the convergence of $\prod\limits_{n=1}^\infty(1+a_n)$, we can assume that $a_n \to 0$, otherwise the product is trivially divergent. So for all but finitely many terms we have $\lvert a_n\rvert < 1$, and then $\operatorname{Re} 1+a_n > 0$, so we can use the principal branch of the logarithm for these. The finitely many terms of the product where $\operatorname{Re} 1+ a_n \leqslant 0$ don't influence the convergence of the product (only the value it converges to, if it converges), so they can be ignored for the purpose. The terms with $a_n = -1$ must be removed from the product (and corresponding logarithm sum) when considering convergence, the terms with $a_n < -1$ could be retained, with an arbitrary choice of the logarithm for these finitely many terms, but it's simpler to also exclude them (but it would have been better to be explicit about that).