The conjecture on finitely generated multiplicatively divisible commutative semirings

Let $S$ be a finitely generated commutative semiring. The semiring $S$ is multiplicatively divisible if for every $a\in S$ and every $n\in\mathbb{N}$ there exists $b\in S$ such that $a=b^n$.

Multiplicative idempotence conjecture. If $S$ is multiplicatively divisible, then $S$ is multiplicatively idempotent.

The conjecture concerns the interaction between finite generation and multiplicative divisibility. The surrounding discussion notes that multiplicatively idempotent semirings are multiplicatively divisible, while algebraically closed fields of characteristic $0$ provide divisible but non-idempotent examples when finite generation is absent.