Let $fg$ mean the functional composition of functions $f, g$.
Let $M$ be a countably infinite, commutative, multiplicative monoid and let $f : M \to (\Bbb{N}_{\gt 0}, \cdot)$ be an isomorphism. Define an ordering on $M$ to be $x \lt y \iff f(x) \lt f(y)$. Then $\lt$ is a total ordering on $M$. In particular for $n,m \in \Bbb{N}$, $n \lt m \iff f^{-1}(m) \lt f^{-1}(n)$, since the latter is $\iff ff^{-1}(m) \lt ff^{-1}(n) \iff m \lt n$.
Proof: if $x \neq y$ then $f(x) \neq f(y)$ and either $f(x) \lt f(y)$ or vv.
Also, $\lt$ is an admissible ordering on $M$.
Proof: we want to show that for all $x, y, z \in M$, if $x \lt y$ then $z x \lt zy$ and $x \lt xy$ except at $y = 1$. This follows from our definition and operations on inequalities in $\Bbb{N}_{\gt 0}$.
Basic Property 1. If $f, g$ both induce $\lt$ on $M$ then $f = g$.
Proof: Let $x \in M$.
$$ f(x) \lt g(x) \iff g^{-1}f(x) \lt x \iff (g^{-1}f)^2(x) \lt g^{-1}f(x) \\ \iff (g^{-1}f)^k(x) \lt (g^{-1}f)^{k-1}(x), \ \forall k \geq 1 $$
This implies there's an infinite strictly descending chain in $M$ given by iterates of $g^{-1}f(x)$, namely $x = (g^{-1}f)^0(x) \gt (g^{-1}f)^1(x) \gt (g^{-1}f)^{2}(x) \gt \dots$ Well we already have by def. that $x \lt y$ in $M$ iff $f(x) \lt f(y)$ in $\Bbb{N}_{\gt 0}$, thus this infinite, upper-bounded, strictly descending chain must also occur in $\Bbb{N}_{\gt 0}$, which is impossible.
Basic Property 2. Every admissible ordering on $M$ is induced by some such bijection.
Is it true and how do I prove / disprove it?
Switch to free monoid on a countable set.
Define $f(1) = 1$. Note that $1$ is automatically the least element in the admissible order on $M$ since $1 \lt 1\cdot x = x, \ \forall x \in M$. Now write generator elements down in their total order: $1 = x_0 \lt x_1 \lt x_2 \lt \dots$ Now define $f(x_i) = p_i$ the $i$th prime, and extend it homomorphically to the rest of $M$. Now $f(\prod_{n=1}^{\infty} x_n^{e_n} \cdot \prod_{n=1}^{\infty} x_n^{d_n}) = f( \prod_{n=1}^{\infty} x_n^{e_n}) \cdot f(\prod_{n=1}^{\infty} x_n^{d_n})$ and it's clearly bijective, so an isomorphism. The admissible order that it induces is that of $M$.
Proof. Define $x \lt' y$ iff $f(x) \lt f(y)$. But for $x = \prod x_i^{e_i}, \ y = \prod y_i^{d_i}$ we have. I give up