**Edit** (following the clarifications of the OP): This is not an answer, it is rather a long comment.

Studying this kind of questions is part of the mission of _factorization theory_: The language of the theory (at least in its classical incarnation) is the language of commutative (and, to a very large extent, [cancellative][1]) [monoids][2], which seems to fit with the OP's idea of a monoid embedding $f$ from the multiplicative monoid of the positive integers into a larger commutative monoid $S$ such that $f(p)$ is a product of atoms${}^{\text{(a)}}$ in an essentially unique way. Note that, in fact, we may assume without loss of generality that $S$ is an atomic monoid (i.e., every non-unit of $S$ factors as a product of atoms).

Now, it is a basic result in the classical theory of factorization that a commutative cancellative monoid $H$ is a unique factorization monoid (i.e., every non-unit of $H$ has an essentially unique factorization into atoms) if and only if $H$ is atomic and every atom is a prime${}^{\text{(b)}}$, if and only the quotient monoid $H/H^\times$ is a free abelian monoid; for a reference, see Theorem 1.2.9 in 

> A. Geroldinger and F. Halter-Koch, _Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory_, Pure Appl. Math. 278, Chapman & Hall/CRC, 2006). 

This means that, *if $S$ is a cancellative and commutative monoid with the unique factorization property*, then there is nothing new we can hope to learn about the arithmetic (multiplicative) structure of $\mathbb N$ from the (mere) existence of the embedding $f$ — and this is true independently of how condition (3) is interpreted. So, for the question to make sense, we need either to give up the idea that $S$ is a cancellative _and_ unique factorization monoid (which seems to be fine with the OP), or to keep track of the _additive_ structure of $\mathbb N$ and rather look, say, for a _semiring_ embedding of $\mathbb N$ into a larger commutative semiring whose multiplicative monoid satisfies conditions (2) and (3). (Honestly, I think condition (3) is still a bit too vague for any sensible answer to be possible.)



• **Notes.** 

(a) An _atom_ of a monoid $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$.  

(b) A *prime* of a monoid $H$ is a non-unit $p \in H$ such that, if $p \mid_H xy$ for some $x, y \in H$, then $p \mid_H x$ or $p \mid_H y$ (here $\mid_H$ is the *divisibility preorder* on $H$, so $a \mid_H b$, for some $a, b \in H$, if and only if $b \in HaH$).


  [1]: https://en.wikipedia.org/wiki/Cancellative_semigroup
  [2]: https://en.wikipedia.org/wiki/Monoid
  [3]: https://en.wikipedia.org/wiki/Free_monoid#The_free_commutative_monoid