$\DeclareMathOperator\Mon{Mon}\DeclareMathOperator\Grp{Grp}$Let $\Mon'$ be the set of isomorphism classes of (small) commutative, unital, cancellative ($a + t = b + t$ implies $a = b$) monoids. It is a commutative unital monoid under product $(M, N) \mapsto M \times N$. The monoid $\Mon'$ is not cancellative because of the ``Eilenberg swindle'' $\prod^\infty \mathbb N \times \mathbb N \simeq \prod^\infty \mathbb N$. For similar reasons, $\mathbb R^2 \simeq \mathbb R$ gives another counterexample.
Restrict to finitely generated monoids $\Mon \subseteq \Mon'$.
Q: Is $\Mon$ a cancellative monoid?
There is a similar monoid of finitely generated abelian groups under cartesian product $\Grp$ which is cancellative by the fundamental theorem of finitely generated abelian groups. The map $\Mon \to \Grp$ which sends a monoid $M$ to its associated group $M^{\mathrm{gp}}$ is monoidal and surjective, so $\Mon$ is not the zero monoid as in the non-finitely generated case. But this is hardly enough to conclude that $\Mon$ is cancellative.
If the answer to the question above is no, there are weaker versions given by demanding that the monoids be ``saturated'' or that the associated group $M^{\mathrm{gp}}$ is torsion free. There are also relative variants for maps either to or from a fixed base monoid $P$, where the multiplication is coproduct/product. Can properties of this monoid tell us about the properties of $P$?
