It's been a while since I've worked with amenable groups and integrals against finitely additive measures, so I could be missing something, but it now seems to me that the question is easy. While apparently in general the direct product of amenable groups isn't amenable, the direct product of abelian groups is of course amenable. So we can use:
Theorem. If $G$ is the direct product of the groups $(G_i)_{i\in I}$ and $G$ is amenable, then there is a finitely additive invariant measure on $G$ that has independence for finite sequences of events depending respectively on pairwise disjoint sets of coordinates.
We need only prove the Theorem for finite sets $I$. For let $F$ be the set of finite partitions of $I$, with a partial order given by $K\le L$ iff $L$ is at least as fine as $K$. For any partition $K\in F$, we consider $G$ to be the product of the $G^A=\prod_{i\in A} G_i$ as $A$ ranges over $K$. The measure $\mu_K$ we get from the finite index-set case of the theorem applied to the product of the $G^A$ will satisfy the independence condition for finite sequences of events depending on pairwise disjoint sets of coordinates that are in the algebra generated by $K$.
Let $\mu$ be a limit point of $\mu_K$ along the net $(F,\le)$. Then $\mu$ is the requisite measure on $G$.
The finite index-set case follows by iterating:
Lemma: If $G_1$ and $G_2$ are groups with finitely additive invariant measures $\mu_1$ and $\mu_2$, then there is a finitely addditive invariant measure $\mu$ on their direct product such that $\mu(A\times B)=\mu_1(A)\mu_2(B)$.
Proof of Lemma: Let $\mu(U) = \int_{G_2} \mu_1 ( \{ x_2 : (x_1,x_2) \in U \} ) d\mu_2(x_2)$. Integrals against f.a. measures are additive so $\mu$ is a f.a. measure, and invariance is obvious.