A monoid is *invertible-free* if $xy=1$ implies $x=y=1$ for all $x,y$.

Question: Can every cancellative invertible-free monoid be embedded in a group?

I'm fairly sure that a quotient of the free product of such a monoid with its mirror (this is the monoid with the same elements and identity but reversed multiplication, i.e. $x\cdot y=yx$) is the "most general" group in which it can be embedded.

This is the non-commutative version of the construction of the integers from the natural numbers.

Does this appear anywhere in the literature as a problem / proposition / theorem?