Let us assume that $S \subset\mathbb{Z}[x]$ is non-empty. This set $S$ has the property that $\mathbb{Z}[x] S \subset S$. However, $S$ is not an ideal because sum of two elements from $S$ is not guaranteed to be in $S$.
$S$ is actually a semigroup in $\mathbb{Z}[x]$, which amounts to saying that $S S \subset S$, which is weaker than $\mathbb{Z}[x] S \subset S$. I have not studied much about semi-groups; hence this question below:
From what I could find in literature and some searching, one could study elasticity of this semigroup, which is a constant related to the factorization of elements of $S$ into irreducibles of $S$. What other properties of semigroups could one study, given that this is a semigroup of polynomials? If one could direct me to a reference that studies semigroups of polynomials, I would be grateful. Thanks.
As @yesterday pointed out, your set $S$ is an ideal of the multiplicative monoid ${\Bbb Z}[X]$.
It is not too difficult to describe the algebraic structure of this monoid. First observe that the polynomial $0$ is a zero of ${\Bbb Z}[X]$ and that $M = {\Bbb Z}[X] \setminus \{0\}$ is a submonoid of ${\Bbb Z}[X]$.
For each non-zero polynomial $p$, let $c(p)$ denote its coefficient of higher degree. For instance, $$ c(-1 -2X + 3X^3) = 3 \quad\text{and}\quad c(1 + X - X^2) = -1 $$ Let $P$ be the set of polynomials $p$ such that $c(p)$ is positive. Let also $I$ be the set of irreducible polynomials of $P$. Since ${\Bbb Z}[X]$ is factorial, $P$ is the free commutative monoid generated by $I$, and since $I$ is countable, it is isomorphic to the additive monoid $({\Bbb N}[X],+)$.
On the other hand, the set of units of ${\Bbb Z}[X]$ is the cyclic group $U = \{-1, + 1\}$. Finally observe that if $p$ is a non-zero polynomial, then either $p \in P$ or $-p \in P$, depending on the sign of $c(p)$. I let you verify that in fact, the monoid $M$ is isomorphic to $U \times P$, the product of the monoids $U$ and $P$.
Altogether, ${\Bbb Z}[X]$ is isomorphic to $(U \times ({\Bbb N}[X],+))^0$, the monoid obtained by adjoining a zero to the monoid $U \times ({\Bbb N}[X],+)$.