Have a look at Grillet's Commutative Semigroups. Let $C$ be a commutative semigroup. The outline of the structure theory is as follows:
As arsmath says, $C$ decomposes as a semilattice of archimedean semigroups. The relevant semilattice is the universal semilattice $C_S = C / (2=1)$ on $C$. The decomposition is as follows: the the fibers of the universal map $C \to C_S$ are archimedean semigroups, called the archimedean components of $C$.
If $C$ is finitely-generated, then so is $C_S$, and hence $C_S$ is finite.
$C$ is said to be complete if each archimedean component contains an idempotent, and subcomplete if each archimedean component embeds into a complete archimedean semigroup. An archimedean semigroup always contains at most one idempotent, so $C$ is complete iff the composite map $C^S \to C \to C_S$ (which is always injective) is an isomorphism. Here $C^S = \{x \in C \mid 2x=x\}$ is the co-universal semilattice on $C$, i.e. the semilattice of idempotents in $C$.
If $C$ is finitely-generated, then $C$ is subcomplete, and its archimedean components are finitely-generated. Thus, arsmath's point that general archimedean commutative semigroups are complicated notwithstanding, for the finitely-generated case, we can focus on the more tractable class of subcomplete archimedean commutative semigroups.
If $C$ is complete archimedean, then it decomposes as an ideal extension $G \to C \to N$ where $G$ is a group and $N$ is nilpotent, i.e. $N$ has an absorbing element $\infty$ such that for every $x\in N$, there is $n \in \mathbb N$ such that $mx = \infty$ for all $m \geq n$ and all $x \in N$. If $C$ is subcomplete archimedean, then it has a similar decomposition where $G$ is cancellative.
If $C$ is finitely-generated and archimedean, then in the decomposition $G \to C \to N$, $G$ is finitely-generated, but $N$ in general is not. Finitely-generated cancellative commutative semigroups are well-understood (they are products of finite groups and cones in $\mathbb Z^n$), so that part of the structure is comprehensible. But this nilpotent part is more mysterious, I think.
Putting it all together, if you have a finitely-generated commutative semigroup $C$, then you can think of it as a finite lattice of a bunch of finite abelian groups equipped with certain positive cones, each of which acts on some nilpotent commutative semigroup, with homomorphisms between these corresponding to the relations in the lattice.