In a class $X$ of algebraic structures, what is the rigorous definition of an $X$-able subset of a member of that class?

Question

There is a subtle distinction in abstract algebra that is often overlooked. Let me illustrate. Consider the monoid $(\mathbb{N};\cdot,1)$ where $\mathbb{N}$ denotes the set of nonnegative integers, and $\cdot$ denotes multiplication. Now, the set $\{0\}$ is not a submonoid of that structure, because it does not contain $1$. However, it is "monoid-able", because it is closed under multiplication and has an identity element. So, my question is basically this. In a class $X$ of algebraic structures, and a member $A$ of the class $X$, what is the rigorous definition of a subset $S$ of $A$ being $X$-able?

Answer 1

The key point allowing the abuse of language you ask about to be understandable is that a monoid is determined by its semigroup reduct (= forget the identity symbol). Conversely, some but not all semigroups can be expanded (= add a symbol referring to an already-existing element) to monoids; additionally, if $M$ is a monoid and $S$ is the semigroup reduct of $M$, not all sub-semigroups of $S$ are expandable to monoids. Finally, even when a subsemigroup $S'$ of $S$ is expandable to a monoid, it may not be expandable to a submonoid of $M$.

That last point is exactly what's going on here (conflating sets and structures for simplicity): $\{0\}$ is a subsemigroup of the semigroup-reduct of $\mathbb{N}$, which is itself expandable to a monoid but is not the semigroup reduct of any submonoid of $\mathbb{N}$.

Sadly, I don't think there is a brief way of describing this situation.