As a complement to my previous non-answer, let me show that the OP's conjecture holds for any order-$3$ semigroup $S$ that is not a group.
Up to the canonical anti-isomorphism between $S$ and its opposite semigroup, there are $17$ such semigroups (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups here. Below, I'll list these $11$ commutative semigorups one by one and prove that they all satisfy the OP's conjecture. It may be a tediousv 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 ordered triple $(x,y,z)$ such that $ac = bc$.
- $S$ 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.
- $S$ 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.
- $S$ 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.
- $S$ 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. However, $yx = zx = x$.
- $S$ 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.
- $S$ is the null semigroup (with zero element $z$) defined by the following table (a null semigroup 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.
- $S$ 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. However, $yx = zx = x$.
- $S$ 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. However, $xy = zy = y$.
- $S$ 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. However, $xz = yz = z$.
- $S$ is the non-unital semilattice defined by the following table (a semilattice 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. However, $xz = yz = z$.
- $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table (a semilattice 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. However, $xz = yz = z$. []
I hope I haven't made too many typos.