Let a (*rational*) *Puiseux monoid* be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (monoid) homomorphism between Puiseux monoids, then there exists a rational number $r \ge 0$ such that $f(x) = r x$ for every $x \in H$.

*Proof.* Set $H^+ := H \setminus \{0\}$ and pick $p, q \in H^+$ (since $f$ maps $0$ to $0$, we can focus on the positive elements). There then exist positive integers $a$, $b$, $c$, and $d$ such that $p = a/b$ and $q = m/n$. It follows that
$$
mb f(p) = f(mbp) = f(ma) = f(naq) = na f(q),
$$
and hence $f(p)/p = f(q)/q$. That is, the function $H^+ \to \mathbb Q_{\ge 0} \colon x \mapsto f(x)/x$ is constant. []

Question.The result, along with a slighly more complicated proof, appears as Proposition 3.2(1) in [F. Gotti, Puiseux monoids and transfer homomorphisms, J. Algebra 516 (2018), 95–114], but I feel it must be older. Do you know of any reference that supports my feeling?

Note that the case of numerical monoids (that is, submonoids of $(\mathbb N, +)$ with finite complement in $\mathbb N$) is covered by Theorem 3.2 in [J.C. Higgins, Representing $N$-semigroups, Bull. Austral. Math. Soc. 1 (1969), 115–125], as I learned from Benjamin Steinberg's answer in another thread some time ago.