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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) monoids, 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 (honestly, I think condition (3) is still a bit too vague for a sensible answer to be possible). 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 condition (2).

• 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 $b \in HaH$).

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) monoids, 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 $b \in HaH$).

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) monoids, 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 would be 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 condition (2).

• 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 $b \in HaH$).

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) monoids, 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 (honestly, I think condition (3) is still a bit too vague for a sensible answer to be possible). 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 condition (2).

• 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 $b \in HaH$).

The following is a slightly edited version of an excerpt from the OP (I'll assume $0 \in \mathbb N$, which doesn't seem to be the case with the OP, for otherwise it's not true that any natural number is a product of primes in an essentially unique way):

Is there any work concerning the existence of an algebraic system $S$ such that (1) $S$ contains the natural numbers $\mathbb N$, (2) the multiplication of integers is extended to this system and is still associative and commutative; (2') every non-zero natural number is a unique product of a set $E$ of indecomposable elements; (3) the structure of $E$ is simpler than the structure of all primes. Addition is not considered yet, so the system we are searching for need not be a ring.

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) monoids, which seems to fit with the OP's idea of an "algebraic system" that extends the multiplication of $\mathbb N$ in such a way that the resulting operation "is still associative and commutative".

With this said, I'm not sure how condition (3) should be interpreted, and I may also have misunderstood the intended meaning${}^{\text{(b)}}$ of condition (2'). But what about the free abelian monoid $\mathscr F_{\rm ab}(X)$ on a given set $X$?

It is easily checked that $\mathscr F_{\rm ab}(X)$ is a commutative, cancellative monoid where every element has an essentially unique factorization into primes${}^{(\text{a})}$ (in particular, the primes of $\mathscr F_{\rm ab}(X)$ are the elements of the basis $X$); in consequence, $\mathscr F_{\rm ab}(X)$ satisfies condition (2). On the other hand, $\mathscr F_{\rm ab}(X)$ contains (a copy of) $X$; so, if $\mathbb N$ is contained in $X$, then (a copy of) $\mathbb N$ is contained in $\mathscr F_{\rm ab}(X)$ and hence condition (1) is also satisfied. I leave it to the OP to decide whether the "structure of $X$" (that is, of the set of primes of $\mathscr F_{\rm ab}(X)$) is simpler than the "structure of the primes of $\mathbb N$".

Incidentally, if the intended meaning of condition (2') is that every element of $S$ (rather than "any natural number") factors in an essentially unique way as a product of "indecomposable elements" (where I take an indecomposable element to be an atom${}^{\text{(b)}}$ in the sense of P.M. Cohn), then 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 the quotient monoid $H/H^\times$ is a free abelian monoid (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).

(a) We let a prime of a monoid $H$ be 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 $b \in HaH$).

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

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) monoids, 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 would be 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 condition (2).

(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 $b \in HaH$).