I've never seen the term 'normal' being used in the way of the OP. A semigroup $S$ (written multiplicatively) with the property that $xS = Sx$ for all $x \in S$ is commonly called either duo or normalizing (see also [this answer][1]). Cancellative, non-commutative examples abound 'in nature'. Here is a short list:

1) Any group (either commutative or not) is a cancellative, duo semigroup.

2) If $(G, \preceq)$ is a totally ordered group (either commutative or not) with identity element $1_G$, then $G^+ := \{x \in G \colon 1_G \prec x\}$ is a cancellative, duo subsemigroup of $G$. For the duoness part, it suffices to note that, for all $a, b \in G^+$, we have $1_G = aa^{-1} \prec aba^{-1} \in G^+$ and, in a similar way, $a^{-1}ba \in G^+$. (This example is already mentioned by user46855 in [their answer][2].)

3) The multiplicative monoid $R^\bullet$ of the non-zero elements of a left (or right) discrete valutation domain $R$ is a strongly Archimedean, cancellative, duo semigroup ('strongly Archimedean' means that, for every $a \in R^\bullet$, there exists an integer $n \ge 1$ such that any product of any $n$ non-zero non-units of $R$ is divisible by $a$ in $R^\bullet$). One special case is given by the ring of formal power series in one variable $x$ over a skew field $D$, with multiplication twisted by a ring automorphism $\sigma$ of $D$ in such a way that
$ax = x\sigma(a)$ for every $a \in D$; see Exercise 19.7 in the 2003 edition of Lam's *Exercises in Classical Ring Theory*. (This example is also already mentioned by user46855 in [their answer][2].)

4) Any direct product of cancellative, duo semigroups is itself cancellative and duo, and it is commutative if and only if each factor in the direct product is (cf. [Anton Klyachko's answer][3]).


  [1]: https://mathoverflow.net/questions/415718/a-name-for-semigroups-in-which-left-and-right-principal-ideals-coincide/416640#416640
  [2]: https://mathoverflow.net/a/160741/16537
  [3]: https://mathoverflow.net/a/155280/16537