Does the convergence of a product imply the convergence of a sum?
$$\prod_{n=0}^\infty(1+a_n)<M_1\implies\sum_{n=0}^\infty a_n<M_2$$
I don't know any properties determinable from the fact that the product converges to say anything about the sum.
If so, what about the other way around?
This implication is not true. Let $a_{2n}=\frac{1}{\sqrt{n}}, a_{2n+1}=\frac{-1}{\sqrt{n}+1}$. Focusing on consecutive terms, we see $a_{2n}+a_{2n+1}=\frac{1}{\sqrt{n}}+\frac{-1}{\sqrt{n}+1}=\frac{1}{n+\sqrt{n}}$ and $(1+a_{2n})\cdot (1+a_{2n+1})=(1+\frac{1}{\sqrt{n}})(1+\frac{-1}{\sqrt{n}+1})=1$. Then clearly the relevant infinite product tends to $1$, but the series diverges.
A simple implication that is true is the following: $\prod_{n=0}^{\infty} (1+|a_{n}|)$ converges $\Rightarrow$ $\sum_{n=0}^{\infty}|a_{n}|$ converges $\Rightarrow$ $\sum_{n=0}^{\infty}a_{n}$ converges because it converges absolutely. (This follows immediately from expanding the terms in the partial products.)
More generally, a well-known result is the following: $\prod_{n=0}^{\infty}(1+a_{n})$ converges absolutely if and only if $\sum_{n=0}^{\infty}a_{n}$ converges absolutely. This is Theorem $6$ of Chapter $5$ in Ahlfors' classic complex analysis text.