Let $S = \{(a,b) : \ a, b \in \Bbb{Z} \wedge \gcd(a,b) \neq 1 \}$. Then it forms a semigroup under componentwise multiplication and if we add an exception, that even though $\gcd(1,1) = 1$, we include $(1,1)$ in set, then we have a semigroup with identity.
The operation $(a,b)\cdot(c,d) \mapsto (ac, bd), \ \ S \times S \to S$. Can be generalized. What if you let the operation be given by a pair of polynomials as $(p(a,b,c,d), q(a,b,c,d))$. For instance the polynomials $p = ac - bd, \ q = ad + bc$ work, i.e. $S$ is closed under $x \cdot y \mapsto (p(x,y), q(x,y))$.
In the following, let $(a,b) = x$ and $(c,d) = y$ and $p(x,y)$ stand for $p(a,b,c,d)$.
Lets see. The polynomials we speak of are in $\Bbb{Z}[x_1, \dots, x_4]$ which we'll just call $R$. Consider the ideal $I = x_1 x_3 R + x_1 x_4 R + x_2 x_3 R + x_2 x_4 R$.
Conjecture: the set of all integer polynomial pairs $p,q$ such that $S$ is closed under $x\cdot y \mapsto (p(x,y), q(x,y))$ is $I \times I$.
Claim $(1)$: For any $p, q \in I$, the pair $p,q$ is such that $S$ is closed under $x \cdot y \mapsto (p(x, y), q(x,y))$
Proof. $\gcd(a,b)\cdot\gcd(c,d) = r$ divides each of $ac, ad, bc, bd$, so when we evaluate each of $p$, $q$, if they're in that ideal then clearly each results in an integer that is divisible by $r$. QED
Claim $(2)$: The set $T$ of all non-constant polynomials $p$ such that $p, q$ is such a pair for some polynomial $q$ forms an ideal of $R$ that contains $I$.
Proof. We've shown that $I$ satisfies the required properties so any such ideal contains $I$. We need to show that for all $p,q \in T, p - q \in T$ and $\forall r \in R, \ rT \subset T$.
Let $r\in R, p \in T$. Then $\exists q \in R$ such that $p, q$ is such a pair. Then $rp, q$ is automatically such a pair as anything dividing $p$ also divides $rp$. Thus $T$ absorbs $R$.
It's tougher to show that $p - q \in T$. We know that $\gcd(p(x,y), q'(x,y)) \neq 1, \forall x, y \in S$ for some non-constant $q' \in R$.