Is the set of all uniquely invertible elements of a semigroup an inverse subsemigroup?

Question

For a semigroup $S$ and $x\in S$, an element $y\in S$ is called an inverse of $x$ iff $xyx=x$ and $yxy=y$. $S$ is called an inverse semigroup when every element of $S$ has a unique inverse. Every group is an inverse semigroup.

Let $S$ be a semigroup, and let $U$ be the set of all elements of $S$ which have unique inverses in $S$. What can we say about this set when $S$ is not an inverse semigroup? In particular, is $U$ a subsemigroup? For $u\in U$, must the unique inverse of $u$ also be in $U?$ Under what conditions are the answers "yes"?

I'm asking this because it seems to be a natural way of generalizing the group of units of a monoids, just like inverse semigroup generalize groups. After some thinking I've decided it didn't seem likely that the answers to the above questions are "yes", which would mean that this generalization isn't good at all. But I haven't been able to prove it.

Supposing I'm right, and this is not a good concept, is there another, better way to generalize the group of units in the inverse semigroup diretion?

Answer 1

The answer to the question:

What can we say about this set when S is not an inverse semigroup? In particular, is U a subsemigroup?

is no. For example, take a semigroup $S$ with $4$ elements $\{a,b,c,d\}$ with multiplication table:

\begin{equation*} \begin{array}{l|llll} &a&b&c&d\\\hline a&a&c&c&d\\ b&c&b&c&d\\ c&c&c&c&d\\ d&c&c&c&d \end{array} \end{equation*}

This is a $2$-element semigroup of right zeros $\{c,d\}$ with two trivial groups $\{a\}$ and $\{b\}$ adjoined. So, the uniquely invertible elements of $S$ are $a$ and $b$, but $ab=c$ and $c$ is not uniquely invertible ($cdc=c$, $dcd=d$ and $c^3=c$).