Let $L=\\{0,1\\}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:M\to L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective. If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.