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. In fact:
Theorem. If the $(G_i)_{i\in I}$ are amenable groups, there is a finitely additive invariant measure on their direct product $G$ that has independence for finite sequences of events depending respectively on pairwise disjoint sets of coordinates.
We need only prove the Theorem for finite $I$. 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 $K\in F$, by the finite index-set case of the theorem, and using the fact that direct products of amenable groups are amenable, choose an invariant measure $\mu_K$ on $G$ that satisfies the independence condition for finite sequences of events depending on pairwise disjoint sets of coordinates that are in the algebra generated by $K$. Choose an ultrafilter on $F$, and let $\mu$ be the pointwise limit of $\mu_K$ along the ultrafilter. Then $\mu$ is the requisite measure on $G$.
But the finite case follows by iterating the application of:
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.