Comparability with identity of an ordered semigroup It is possible to compare any ordered semigroup with $0$:
Comparability with zero of an ordered semigroup
Let's say an ordered semigroup $S$ is comparable with identity if it can be embedded into an ordered monoid $S^1$ in which every element is comparable with $1$, and the order of $S$ is a subset of the order of $S^1$.
Clearly, any totally ordered monoid is a semigroup comparable with $1$.
An ordered group $G(\bullet, e, \le)$ is comparable with identity if and only if it is totally ordered:

*

*$a \le b \iff e \le a^{-1} \bullet b$;

*$a \le b \iff e \le b \bullet a^{-1}$.

There is an example of a totally ordered semigroup that is not comparable with $1$: https://math.stackexchange.com/a/3739556/427611
What would be the conditions for an ordered semigroup to be comparable with identity?
 A: As there is no answer so far, here is a small observation related to your question.
Let $M$ be an ordered monoid and suppose that, for each $x \in M$, $x \leqslant 1$. Note that this is a stronger condition that just requiring that every element is comparable with $1$. I claim that in this case, $M$ is ${\cal J}$-trivial, which means that the Green's relation ${\cal J}$ is the equality on $M$.
Suppose that $x \leqslant_\mathrel{\cal J} y$. By definition, there exists $u, v \in M$ such that $x = uyv$. Since $u \leqslant 1$ and $v \leqslant 1$, the condition  $x = uyv$ implies $x \leqslant y$.
Now, if $x \mathrel{\cal J} y$, then by definition $x \leqslant_\mathrel{\cal J} y$ and $y \leqslant_\mathrel{\cal J} x$, whence $x \leqslant y$ and $y \leqslant x$ and finally $x = y$.
Since a subsemigroup of a ${\cal J}$-trivial monoid is also ${\cal J}$-trivial, it follows that a semigroup which is not ${\cal J}$-trivial cannot be embedded in a monoid $M$ in which $x \leqslant 1$ for all $x \in M$.
A (difficult) result states that a finite monoid is ${\cal J}$-trivial if and only if it is a quotient of a finite monoid $M$ in which $x \leqslant 1$ for all $x \in M$, but unfortunately, I have no such result to offer for submonoids.
