I should read J. C. Rosales and P. A. García-Sánchez's book Finitely Generated Commutative Monoids and L. Redei's book The Theory of Finitely Generated Commutative Semigroups. I haven't. But here's what I've heard so far:
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If a commutative monoid is finitely generated it is finitely presented.
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If a finitely generated commutative monoid is cancellative ($a + b = a' + b \Rightarrow a = a'$) then it embeds in a finitely generated abelian group.
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If a finitely generated commutative monoid is cancellative and torsion-free (for any natural number $n \ge 1,$ $n a = n b \Rightarrow a = b$) then it embeds in a finitely generated free abelian group. (This follows easily from the previous claim.)
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If a commutative monoid is a submonoid of $(\mathbb{N},+,0)$ it is called a numerical monoid and of course it is cancellative.
A lot is known about numerical monoids, though I don't believe they have been "classified" in any useful sense.
If we drop the property of being cancellative we get an enormous wilderness of finitely generated commutative monoids, so there shouldn't be any simple 'classification theorem'. But there still might be interesting structure theorems which help us understand this wilderness, just as there are for (say) compact topological abelian groups. What are they?
Best Answer
The comments are getting a bit long so I'll put this as a partial answer. The case of von Neumann regular commutative semigroups was handled by Clifford in the 1940s. A semigroup is von Neumann regular if for all $a$, there exists $b$ with $aba=a$. Clifford proved a regular commutative semigroup is the same thing as a pair $(E,F)$ where $E$ is a poset with binary meets and $F$ is a presheaf of abelian groups on $E$. If the semigroup is a finitely generated monoid then $E$ will be a finite lattice.
For example, given such a pair, the underlying set of the semigroup is the disjoint union of the $F(e)$ with $e$ in $E$ (so the arrow set of the associated discrete fibration). The product of $a \in F(e)$ with $b \in F(e')$ is obtained by restricting both elements to the meet of $e$ and $e'$ and taking their product.
The more general semilattice decompositions in the comments are not as good as this.