Let $M = \Bbb{N}^{\times}$ and $N = (\Bbb{N}\setminus \{1\} \cup \{0\})^+$ be the multiplicative, and respectively, the additive naturals and let $\varphi : M \to N$ be defined by taking $1$ to $0$ and $\prod_{i} p_i$ to $\sum_i p_i$ for any product of primes of the primes $p_i$. Then we're clearly looking at a well-defined monoid homomorphism.
Consider the kerenel pair $\ker \varphi = \{ (x,y) \in M \times M : \varphi(x) = \varphi(y)\}$. It defines a congruence relation on $M$ and so we can take the quotient:
$$ M' = M/\ker \varphi $$
First note that stuff happens such as $\varphi(39) = \varphi(55)$ since $3 + 13 = 16 = 5 + 11$ so that $\ker \varphi$ is indeed non-trivial.
The wikipedia article states:
It turns out that $\ker f$ is an equivalence relation on $M$, and in fact a congruence relation. Thus, it makes sense to speak of the quotient monoid $M/(\ker f)$. The first isomorphism theorem for monoids states that this quotient monoid is naturally isomorphic to the image of $f$ (which is a submonoid of $N$),(for the congruence relation).
Thus $M' \simeq N$, since $\varphi$ is surjective. How can we easily show that $\varphi$ is surjective? It seems like maybe we can use $(p) + (q) = \Bbb{Z}$ for two prime ideals $p,q$ but we can't really since negative integers would be involved.
If $n$ is even, $n = \varphi(2^{n/2})$ and if $n$ is odd, $n = \varphi(2^{(n-3)/2}\times 3)$.