This question is motivated by a partial answer to another question on MO.
Given an infinite finitely generated group $G$ and a finite generating set $S$, let $b_n^S$ be the cardinality of the ball of radius $n$ [around the identity].
Say a pair $G$ has pinched exponential growth w.r.t. $S$ if there are constants $K<L \in \mathbb{R}_{> 0}$ and $M \in \mathbb{R}_{>0}$ so that $K \cdot \mathrm{exp}(Mn) \leq b_n^S \leq L \cdot \mathrm{exp}(Mn).$
Question: What are examples of amenable groups $G$ which have exponential growth, but there is no choice of generating set $S$, so that the group has pinched exponential growth (w.r.t. $S$)?
Remarks:
the question could be strengthened to require that the group does not have a rational growth series. (If there are more than one pole on the radius of convergence of the growth series, then the group does not have pinched exponential growth.)
the question could also be strengthened to require that the group be finitely presented.
an equivalent formulation of the question is: let $g(S) =\displaystyle \lim_{n \to \infty} \sqrt[n]{b_n^S}$. Is there an $S$ so that $\limsup_n \dfrac{ b_n^S}{g(S)^n} < \infty$?
[EDIT: Further remarks:
there would be a weaker form of the question where non-amenable groups could be considered. Non-amenable groups are however not interesting in this context (see linked question and comment below)
[complement to a comment below] If $S$ is a generating set, let $S' = S \cup \lbrace e \rbrace$. Assume $G$ has pinched exponential growth w.r.t. $S_G$ and $H$ has pinched growth w.r.t. to $S_H$. Consider the generating sets $S_+ = (S_G \times \lbrace e_H \rbrace) \cup (\lbrace e_G \rbrace \times S_H)$ and $S_\times = S_G' \times S_H'$. Then it's not too difficult to check that $G \times H$ has pinched growth w.r.t. $S_\times$ (because the balls are just product of the balls). It does not have however pinched growth w.r.t. $S_+$. (To see this note that the balls have the form $\cup_{i =0}^n \sigma_i(G) \times \sigma_{n-i}(H)$ where $\sigma_i$ are the sphere of radius $i$. Combined with the fact that the sphere are also pinched [see linked answer for details] gives the fact that the growth is not pinched)
[Still extension of that comment] Assume $G$ has pinched growth w.r.t. to $S_G$ and $H = \mathbb{Z}$ with $S_H = \lbrace 1;-1 \rbrace$ as generating. Then $G \times H$ has pinched growth w.r.t. $S_+$ but not w.r.t. $S_\times$. Indeed, the spheres w.r.t. $S_+$ have cardinality of the balls in $G$. So the cardinality of the balls w.r.t. $S_+$ can be bounded by geometric sums. On the other hand the balls w.r.t. $S_\times$ have the cardinality of a ball in $\mathbb{Z}$ times a ball in $G$ (i.e. bounded above and below by some function of the form $n exp(Mn)$). Hence not pinched. The case of $G=F_2$ is particularly easy to check (see comment below).
If $G$ has pinched growth, then $G \times \mathbb{Z}^2$ does not have pinched growth w.r.t. to either $S_\times$ or $S_+$. But that does not exclude another generating set with pinched growth.
Most "classical" amenable groups of exponential growth have a generating set with pinched growth (this includes the lamplighter group, solvable Baumslag-Solitar groups and metabelian polycyclic groups).
END of EDIT]