As a complement to my [previous non-answer][1], let me show that the OP's conjecture holds for any order-$3$ commutative semigroup $S$ that is not a group (equivalently, that is not a cyclic group of order $3$).
Up to the canonical anti-isomorphism between a semigroup and its [opposite][2], there are $17$ semigroups of order $3$ (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups [here][3]. Below, I'll list these $11$ commutative semigroups one by one and prove that they all satisfy the OP's conjecture. It may be a tedious task, but is going to help with another reduction.
All semigroups will be defined on the set $\{x, y, z\}$ through their Cayley tables. Since $\lfloor (3-1)/2 \rfloor = 1$, we need to find, for each semigroup on the list, a permutation $(a,b,c)$ of the triple $(x,y,z)$ such that $ac = bc$.
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1. $S$ is the cyclic semigroup (with $y = x^2$, $z = x^3$, and $x^2 = x^4$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | y | z | y |
| **y** | z | y | z |
| **z** | y | z | y |
It satisfies the OP's conjecture by Proposition 2 [here][1].
2. $S$ is the cyclic semigroup (with $y = x^2$, $z = x^3$, and $x^3 = x^4$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | y | z | z |
| **y** | z | z | z |
| **z** | z | z | z |
It satisfies the OP's conjecture by Proposition 2 [here][1].
3. $S$ is the monoid (with identity element $z$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | y | x |
| **y** | y | y | y |
| **z** | x | y | z |
It is isomorphic to the (unital) submonoid $\{0, \pm 1\}$ of the integers under multiplication. In particular, it has a non-trivial unit (namely, $x$); hence, it satisfies the OP's conjecture by Proposition 1 [here][1].
4. $S$ is the monoid (with identity element $y$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | x | x |
| **y** | x | y | z |
| **z** | x | z | z |
Its group of units is trivial, so we can't apply any of Propositions 1 to 3 [here][1]. However, $yx = zx = x$.
5. $S$ is the nilsemigroup (with zero element $z$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | x | x |
| **y** | x | z | z |
| **z** | x | z | z |
It satisfies the OP's conjecture by Proposition 3 [here][1].
6. $S$ is the null semigroup (with zero element $z$) defined by the following table (a [_null semigroup_][4] is a semigroup with zero in which any product of any two elements is zero):
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | z | z |
| **y** | z | z | z |
| **z** | z | z | z |
Every null semigroup is a nilsemigroup, so $S$ satisfies the OP's conjecture by Proposition 3 [here][1].
7. $S$ is the non-unital semigroup defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | z | z |
| **y** | z | y | z |
| **z** | z | z | z |
It is neither a cyclic semigroup nor a nilsemigroup ($y$ and $z$ are both idempotents), so we can't apply any of Propositions 1 to 3 [here][1]. However, $yx = zx = x$.
8. $S$ is the non-unital semigroup defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | y | z |
| **y** | y | y | y |
| **z** | z | y | z |
It is neither a cyclic semigroup nor a nilsemigroup ($y$ and $z$ are both idempotents), so we can't apply any of Propositions 1 to 3 [here][1]. However, $xy = zy = y$.
9. $S$ is the non-unital semigroup defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | z | x | z |
| **y** | x | y | z |
| **z** | z | z | z |
It is neither a cyclic semigroup nor a nilsemigroup ($y$ and $z$ are both idempotents), so we can't apply any of Propositions 1 to 3 [here][1]. However, $xz = yz = z$.
10. $S$ is the non-unital semilattice defined by the following table (a [_semilattice_][5] is a commutative semigroup in which every element is idempotent):
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | x | y | z |
| **y** | y | y | z |
| **z** | z | z | z |
It is neither a cyclic semigroup nor a nilsemigroup, so we can't apply any of Propositions 1 to 3 [here][1]. However, $xz = yz = z$.
11. $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table:
| | x | y | z |
| :-: |:-:|:-:|:-:|
| **x** | x | y | z |
| **y** | y | y | z |
| **z** | z | z | z |
It is neither a cyclic semigroup nor a nilsemigroup, so we can't apply any of Propositions 1 to 3 [here][1]. However, $xz = yz = z$. []
I hope I haven't made too many typos.
[1]: https://mathoverflow.net/a/480451/16537
[2]: https://en.wikipedia.org/wiki/Opposite_category
[3]: https://en.wikipedia.org/wiki/Semigroup_with_three_elements
[4]: https://en.wikipedia.org/wiki/Null_semigroup
[5]: https://en.wikipedia.org/wiki/Semilattice