This question is motivated by the following well-known theorems:

Thm (Plünnecke):If $A$ is a finite nonempty subset of an abelian group, then for every $n$ we have $|A^n| \le \frac{|AA|^n}{|A|^n}|A|$.

Thm (Ruzsa):If $A$ is a finite nonempty subset of a group, then for every $n$ we have $|A^n| \le \frac{|AAA|^{2n}}{|A|^{2n}}|A|$.

Here by $A^n$ I mean the set of all products of $n$ elements of $A$, so $A^2 = AA$ and $A^3 = AAA$.

I would like to know if there is any similar generalization to cancellative semigroups. Specifically:

Question:Do there exist integers $k, c$ such that for every finite nonempty subset $A$ of any cancellative semigroup and for every $n$, $|A^n| \le \frac{|A^k|^{cn}}{|A|^{cn}}|A|$?

**Edit:** There is a counterexample in the non-cancellative case. For any $n$, let $E_n = \langle e \mid e^{n+2} = e^{n+1}\rangle$. For any group $G$, let $S$ be the quotient of $(G \times E_n)\cup\{0\}$ where we identify $(g,e^{n+1})$ with $0$ for every $g\in G$. Let $A$ be the image of $G\times \{e\}$ in $S$. Then $|A| = |AA| = \cdots = |A^n| = |G|$, but $|A|^{n+k} = 1$ for every $k \ge 1$. Taking a product of many examples like this and a free semigroup, we can arrange that $|A| = |AA| = \cdots = |A^n|$, but $|A^{n+k}| = |A|^{(n+k)/n}$ for every $k \ge 1$.

Here's an easy result which actually uses cancellativity:

Thm:If $A$ is a subset of a cancellative semigroup $S$, then there is a subset $P \subseteq AA$ with $|P| \ge \frac{|A|}{2}$ such that for any subsets $C,B$ of $S$, we have $|CPB| \le 2\frac{|CA|}{|A|}\frac{|AB|}{|A|}|AA|$. In particular, $|AP^nA| \le 2^n\frac{|AA|^{2n}}{|A|^{2n}}|AA|$.

To prove this, take $P$ to be the set of products in $AA$ which can be written as a product in at least $\frac{|A|^2}{2|AA|}$ ways (the "popular" products) and write down a clever injection.