Let $\mathbb A = (A, +\_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ such that $\mathbb A$ is a submonoid of $\mathbb B$ and $x$ is left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below)?


It is known from a paper of A. I. Mal'cev [Math. Ann. **113** (1937), No. 1, 686-691] that there exist (finitely generated) cancellative monoids that do *not* embed into a group, but the question here is about an embeddability condition that looks very much weaker.



Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible
The answer is no. 


In a cancellative monoid left invertible elements are right invertible. If yx =1 then xyx=x and so xy =1 by cancellation. So you are asking to invert a given element in some cancellative monoid. 


Take a fg cancellative monoid M
 not embeddable in a group. Invert the generators one by one to embed M in the group of units of a monoid and get a contradiction. 


