[In a comment to the OP, I asked whether a statement along the lines of Corollary 1 below would count as an "interesting structure theorem", and this post expands on John Baez's yes to that question.]

Let $H$ be a multiplicatively written monoid; unless stated otherwise, $H$ need not be commutative, cancellative, or whatever. An *atom* of $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$. We write $H^\times$ and $\mathscr A(H)$, resp., for the group of units and the set of atoms of $H$; and $\mid_H$ for the *divisibility preorder* on $H$, that is, the binary relation on $H$ defined by $x \mid_H y$ iff $y \in H$ and $x \in HyH$.

Denote by $\mathscr F(X)$ the free monoid on a basis $X$; we'll refer to the elements of $\mathscr F(X)$ as *$X$-words* and write $\|\cdot\|_X$ for the *length* of an $X$-word (see the section "Pills of factorization theory" in this answer if something in the terminology or in the notation doesn't look familiar). Given a non-unit $x \in H$, we define $\mathcal Z_H(x)$ as the set of all $\mathscr A(H)$-words $\mathfrak a$ such that $x = \pi_H(\mathfrak a)$, where $\pi_H$ is the canonical (monoid) epimorphism $\mathscr F(H) \to H$. The $\mathscr A(H)$-words in $\mathcal Z_H(x)$ are the *atomic factorizations* of $x$ (relative to $H$).

We write $\sqsubseteq_H$ for the *shuffling preorder* induced on $\mathscr F(H)$ by $\mid_H$, that is, the binary relation (in fact, a preorder) on the set of $H$-words given by $\mathfrak a \sqsubseteq_H \mathfrak b$ if and only if $\mathfrak a, \mathfrak b \in \mathscr F(H)$ and there is an injective function $\sigma \colon [\![1, \|\mathfrak a\|_H ]\!] \to [\![1, \|\mathfrak b\|_H ]\!]$ such that the $i$-letter of $\mathfrak a$ is $\mid_H$-equivalent${}^{(2)}$ to the $\sigma(i)$-th letter of $\mathfrak b$ for each $i \in [\![1, \|\mathfrak a\|_H ]\!]$. Note that two $H$-words are $\sqsubseteq_H$-equivalent only if they have the same length.

We call $H$ *atomic* (resp., FF-*atomic*) if $\mathcal Z_H(x)$ is non-empty (resp., finite and non-empty up to $\sqsubseteq_H$-equivalence) for every non-unit $x \in H$; and FmF-*atomic* if, for each non-unit $x \in H$, the set of $\sqsubseteq_H$-minimal $\mathscr A(H)$-words in $\mathcal Z_H(x)$ is finite and non-empty up to $\sqsubseteq_H$-equivalence.

Lastly, we say that $H$ is *finitely generated up to units* (or shortly, an *f.g.u. monoid*) if there is a finite subset $A \subseteq H$ such that $H \setminus H^\times$ is contained in the subsemigroup generated by the set $H^\times A H^\times$; and is *locally finitely generated up to units* (or shortly, an *l.f.g.u. monoid*) if, for each $a \in H$, the smallest divisor-closed submonoid${}^{(3)}$ of $H$ containing $a$ is an f.g.u. monoid.

**Examples.** (i) If $H$ is the multiplicative monoid of the ring of integers, then $\mathscr A(H)$ is the set of (positive or negative) primes, and the $H$-words $(-2) \ast 3 \ast (-5)$ and $(-3) \ast 5 \ast 2$ are $\sqsubseteq_H$-equivalent atomic factorizations of $-30$.

The monoid $H$ is not atomic: $0$ is a non-unit of $H$ with no atomic factorizations. Note, though, that $0$ is *irreducible* in the sense of D.D. Anderson and S. Valdes-Leon (see the definition before Theorem 2 below) and hence $H$ has the weaker property of being *factorable* (that is, every non-unit is a product of irreducibles).

(ii) If $p \in \mathbf N^+$ is a prime, $n$ is an integer $\ge 2$, and $H$ is the multiplicative monoid of the ring of integers modulo $p^n$, then the $H$-word $\bar{p}^{\ast (n+1)}$ is an atomic factorization of $\bar{0}$ but is not $\preceq_H$-minimal (here, $\bar{x}$ is the residue class mod $p^n$ of an integer $x$).

In fact, $H$ is FmF-atomic, but not FF-atomic: In particular, if $k$ an integer $\ge n$ and $u_1, \ldots, u_k \in H$ are units, then the $H$-word $\bar{p}u_1 \ast \cdots \bar{p}u_k$ is an atomic factorization of $\bar{0}$, but it is a $\sqsubseteq_H$-minimal atomic factorization if and only if $k = n$.

With all these premises in place, we have:

**Theorem 1.** Every acyclic${}^{(4)}$, l.f.g.u. monoid is FmF-atomic.

Most notably, this implies the following:

**Corollary 1.** If $H$ is a unit-cancellative${}^{(5)}$, commutative monoid and the quotient monoid $H/H^\times$ is finitely generated, then $H$ is FF-atomic.

The corollary is essentially due to A. Geroldinger and F. Halter-Koch, cf. Proposition 2.7.8.4 in their monograph:

*Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory*, Pure Appl. Math. **278**, Chapman & Hall/CRC, Boca Raton (FL), 2006.

These results are basically a generalization of the *existence part* of the fundamental theorem of arithmetic and, at least to some extent, bring about the same kind of "structural content" (though atomic factorizations are, in general, far from being unique in any sensible way). In fact, they are part of a bigger picture; and though the OP asked only about *finitely generated* (commutative) monoids, it is perhaps worth seeing what can be done in a greater generality.

In particular, taking an *irreducible* of $H$ to be a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$ with $HxH \ne HaH \ne HyH$ leads to:

**Theorem 2.** If $H$ is a Dedekind-finite${}^{(6)}$ monoid satisfying the ACCP${}^{(7)}$, then every non-unit of $H$ factors as a product of irreducibles.

To my knowledge, a proof of Theorem 2 for *commutative* rings was first published by D.D. Anderson and S. Valdes-Leon in

*Factorization in Commutative Rings with Zero Divisors*, Rocky Mountain J. Math. **26** (1996), No. 2, 439-480;

their proof (see loc. cit., Theorem 3.2) carries over verbatim to *commutative* monoids (and is, in some sense, straightforward from the definitions). In a different direction, we have:

**Theorem 3.** The conditions below are equivalent:

- $H$ is acyclic and satisfies the ACCP.
- $H$ is unit-cancellative and satisfies the ACCPR and the ACCPL${}^{(8)}$.

Moreover, each of these conditions implies that $H$ is atomic.

Note that every atom is irreducible, and the converse is true if $H$ is acyclic. Insofar as I'm aware, the terms "atom" and "atomic" were first coined by P.M. Cohn, whom I believe deserves credit also for the statement and the proof of the following (although some people do not agree with me on this last point, I haven't yet seen any convincing evidence to claim the contrary):

**Corollary 2.** If $H$ is cancellative and satisfies the ACCPR and the ACCPL, then it is atomic.

There is more to the story. For further details and generalizations, I can only address the interested reader to an article of mine on monoids and preorders (here) and some related papers with Austin A. Antoniou (here) and Laura Cossu (here and here).

**Notes.**

(1) $a \mid_H b$, for some $a, b \in H$, if and only if $b \in HaH$.

(2) Given a preorder $\preceq$ on a set $X$, we say that two elements $x, y \in H$ are *$\preceq$-equivalent* if $x \preceq y \preceq x$.

(3) A submonoid $K$ of $H$ is *divisor-closed* if $x \mid_H y$ for some $x \in H$ and $y \in K$ implies that $x \in K$ (among other things, this guarantees that $H$ and $K$ have the same units).

(4) $H$ is *acyclic* if $xyz \ne y$ for all $x, y, z \in H$ such that either $x$ or $z$ is not a unit.

(5) $H$ is *unit-cancellative* if $xy \ne x \ne yx$ for all $x, y \in H$ such that $y$ is a non-unit. Obviously, every cancellative monoid is unit-cancellative; moreover, every acyclic monoid is unit-cancellative and the two notions coincide on the level of commutative monoids. On the other hand, Adian's embedding theorem (that is, the special case of Guba's embedding theorem where the left and right graphs of a finite semigroup presentation are cycle-free) implies at once the existence of a finitely generated, cancellative monoid with trivial group of units that is not acyclic.

(6) $H$ is *Dedekind-finite* if every left- or right-invertible element is a unit (or equivalently, if $xy = 1_H$ for some $x, y \in H$ implies $yx = 1_H$). It is fairly clear that every unit-cancellative or commutative monoid is Dedekind-finite.

(7) $H$ satisfies the ACCP if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $Ha_nH \subsetneq Ha_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $HaH$ with $a \in H$ is called a *principal ideal* of $H$).

(8) $H$ satisfies the ACCPR if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $a_nH \subsetneq a_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $aH$ with $a \in H$ is called a *principal right ideal* of $H$); and it satisfies the ACCPL if the opposite monoid of $H$ satisfies the ACCPR.

nowmy last sentence, though it wasn't when Andres Caicedo asked. $\endgroup$20more comments