Natural Numbers under Multiplication form Ordered Commutative Semigroup

Theorem
Let $\left({S, \circ, *, \preceq}\right)$ be a naturally ordered semigroup with product.

Then $\left({S, *, \preceq}\right)$ is an ordered commutative semigroup.

Proof
From Multiplication in Naturally Ordered Semigroup is Closed, $*$ is a closed operation on $S$.

From Multiplication in Naturally Ordered Semigroup is Associative, $*$ is an associative operation on $S$.

From Multiplication in Naturally Ordered Semigroup is Commutative $*$ is a commutative operation on $S$.

Thus by definition $\left({S, *}\right)$ is a commutative semigroup.

Let $a, b \in S$ such that $a \preceq b$.

Let $n \in S$.

Let $a = b$.

Then $n * a = n * b$ and so:
 * $n * a \preceq n * b$

and $a * n = b * n$ and so:
 * $a * n \preceq b * n$

Let $a \ne b$.

Then $a \prec b$.

If $n \ne 0$ then from Ordering on Naturally Ordered Semigroup Product:
 * $n * a \prec n * b$

By Multiplication in Naturally Ordered Semigroup is Commutative:
 * $a * n \prec b * n$

and so:
 * $n * a \preceq n * b$
 * $a * n \preceq b * n$

If $n = 0$ then:
 * $n * a = 0 = n * b$
 * $a * n = 0 = b * n$

and so:
 * $n * a \preceq n * b$
 * $a * n \preceq b * n$

Thus in all cases, for all $a, b, n \in S$:
 * $a \preceq b \implies n * a \preceq n * b \land a * n \preceq b * n$

Thus $\preceq$ is seen to be compatible with $*$.

So $\left({S, *, \preceq}\right)$ is seen by definition to be an ordered commutative semigroup.