Here's a partial answer to the Stone space question. The answer is yes for metrizable Stone spaces: if $X \cong X + X + X$ then $X \cong X + X$. I assume you're using $+$ to denote coproduct of topological spaces.
Proof: Write $I(X)$ for the set of isolated points of a topological set $X$. (A point is isolated if, as a singleton subset, it is open.) Then $I(X + Y) \cong I(X) + I(Y)$ for all $X$ and $Y$. So, supposing that $X \cong 3X$, we have $I(X) \cong 3I(X)$. But $X$ is compact, so $I(X)$ is finite, so $I(X)$ is empty. Hence $X$ is a compact, metrizable, totally disconnected space with no isolated points. A classical theorem then implies that $X$ is either empty or homeomorphic to the Cantor set. In either case, $X \cong X + X$.
I guess metrizability of the Stone space corresponds to countability of the corresponding Boolean ring.
The topological theory of Stone spaces is more subtle in the non-metrizable case, if I remember correctly.