The following is a slightly edited version of an excerpt from the OP:
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') any 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 of condition (2'). But what about the free abelian monoid $\mathscr F_{\rm ab}(X)$ over an infinite 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 $\mathbb N$".
• Notes.
(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) 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 irreducible elements of $S$ (where I take an irreducible element to be an atom in the sense of Cohn), then it is a basic result in the classical theorem 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 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).