The answer is no. Consider the family of groups $G_k:=(\mathbf Z_2{}^k)\rtimes \mathbf Z_2$, where the group on the right acts by interchanging the first and second coordinate. Then the commutator subgroup $G_k'$ is generated by $g_k:=((1,1,0,\ldots, 0),0)$, i.e. is of order two. So $G_k$ is non-abelian (of arbitrarily large order), but as little non-abelian as possible. Note that each element in $G_k$ has order dividing 4.

Now let $A=\lbrace a_1,\ldots, a_n \rbrace \subseteq G_k$. Then any element in $A^m$ can be written as $a_1^{e_1} \cdots a_n^{e_n} g_k^{e_k}$, where $0\leq e_i \leq 3$ for $i=1,\ldots n$ and $e_k=0,1$. This follows by applying the identity $ab=[a,b]ba$ repeatedly. In particular, $|A^m|$ is bounded by a constant depending on $n$ only, not on $m$.

The answer is no. Consider the family of groups $G_k:=(\mathbf Z_2{}^k)\rtimes \mathbf Z_2$, where the group on the right acts by interchanging the first and second coordinate. Then the commutator subgroup $G_k'$ is generated by $g_k:=((1,1,0,\ldots, 0),0)$, i.e. is of order two. So $G_k$ is non-abelian (of arbitrarily large order), but as little non-abelian as possible. Note that each element in $G_k$ has order dividing 4.

Now let $A=\lbrace a_1,\ldots, a_n \rbrace \subseteq G_k$. Then any element in $A^m$ can be written as $a_1^{e_1} \cdots a_n^{e_n} g_k^{e_k}$, where $0\leq e_i \leq 3$ for $i=1,\ldots n$ and $e_k=0,1$. This follows by applying the identity $ab=[a,b]ba$ repeatedly. In particular, $|A^m|$ is bounded by a constant depending on $n$ only, not on $m$.

Edit: To be explicit, any two elements in $G_k$ will generate a subgroup of order at most 32, so $(n,m)=(2,31)$ is one counter-example.

The answer is no. Consider the family of groups $G_k:=(\mathbf Z_2{}^k)\rtimes \mathbf Z_2$, where the group on the right acts by interchanging the first and second coordinate. Then the commutator subgroup $G_k'$ is generated by $((1,1,0,\ldots, 0),0)$, i.e. is of order two. So $G_k$ is non-abelian (of arbitrarily large order), but as little non-abelian as possible.

From $ab=[a,b]ba$ it follows that for a subset $A\leq G_k$ of cardinality $n$, $|A^m|\leq 2\cdot \binom{n}{m}$. Choosing $n,m$ such that $2\cdot {n \choose m}\leq \binom{n+m-1}{m}$ implies the claim.

The answer is no. Consider the family of groups $G_k:=(\mathbf Z_2{}^k)\rtimes \mathbf Z_2$, where the group on the right acts by interchanging the first and second coordinate. Then the commutator subgroup $G_k'$ is generated by $g_k:=((1,1,0,\ldots, 0),0)$, i.e. is of order two. So $G_k$ is non-abelian (of arbitrarily large order), but as little non-abelian as possible. Note that each element in $G_k$ has order dividing 4.

Now let $A=\lbrace a_1,\ldots, a_n \rbrace \subseteq G_k$. Then any element in $A^m$ can be written as $a_1^{e_1} \cdots a_n^{e_n} g_k^{e_k}$, where $0\leq e_i \leq 3$ for $i=1,\ldots n$ and $e_k=0,1$. This follows by applying the identity $ab=[a,b]ba$ repeatedly. In particular, $|A^m|$ is bounded by a constant depending on $n$ only, not on $m$.