Does there exist a finite monoid $M$ such that for some $x \in M,$ the following hold?
$x$ cancels on both the left and the right: $$\frac{ax=bx}{a=b}\qquad \frac{xa=xb}{a=b}$$
$x$ has no two-sided inverse:
$$(\forall y \in M)(xy \neq 1 \vee yx \neq 1)$$
I am especially interested in the case where $M$ is commutative.
I'm not a monoids person, so I could be way off base here, but here goes:
If $M$ is commutative, then a left- and right-inverse are the same (i.e., if one exists, so does the other. I suppose that there could be 2 or more of each...)
In that case, fix $x$ and look at the map $$ h : M \to M : u \mapsto xu. $$
If that's NOT injective, i.e., if $xu = xv$ for some $u$ and $v$ with $u \ne v$, then you don't have cancellation. So by your hypotheses, $h$ is injective.
Because $M$ is a finite set, that makes $h$ surjective as well. So there's a value $u \in M$ with $hu = 1$. (Assuming your monoid HAS a unit, of course.)