Let $(R,+,\cdot)$ be a semiring imbued with a regular Hausdorff topology such that addition and multiplication are continuous. Suppose we have functions $f:X\to R$ and $g:Y\to R$ which are unconditionally summable in the sense of $(R,+)$ being an abelian topological group. Does it necessarily follow that their product $(x,y)\mapsto f(x)\cdot g(y)$ is unconditionally summable over the domain $X\times Y$? It should at least be conditionally summable, since continuity of product and the distributive properties gives the following equality. $$\left(\sum_{x\in X} f(x)\right)\cdot \left(\sum_{y\in Y} g(y)\right) = \sum_{x\in X} \sum_{y\in Y} (f(x)\cdot g(y))$$
My question is if the sum $\sum_{(x,y)}f(x)g(y)$ also converges unconditionally. I'm looking either for a proof that the sum always converges unconditionally, or else a counter example where the claim fails. As a few non-examples, the claim holds true when $R=\mathbb{C}$ is the field of complex numbers under the usual topology, since we can easily prove absolute convergence. The claim likewise holds in the extended natural numbers, and more generally in any case where the sums include an annihilating element like $\infty$. The claim should also hold true in any bounded distributive lattice, since conditionally convergent sums always converge unconditionally.
As some points of interest, continuity of product and the distributive property prove that for each fixed $x\in X$, the sum $\sum_{y\in Y}f(x)\cdot g(y)$ converges unconditionally. It can also be shown that a sum of two unconditional sums is again unconditional, so it follows that for any finite subset $S\subseteq X$, the sum $\sum_{(x,y)\in S\times Y} f(x)\cdot g(y)$ converges unconditionally. This suggests that, if a counter example exists, both $X$ and $Y$ must be infinite. On the other hand, an infinite sum of unconditional sums need not converge unconditionally, even if it converges. For example let $X=\{1,-1\}$ and $f:X\to\mathbb{R}$ be the inclusion map, then although $\sum_x f(x)=0$ converges trivially, the sum $\sum_{n\in\mathbb{N}}\sum_x f(x)=0$ only converges conditionally, since we can't swap the order of the sums.