This is not exactly an answer to the question, but is instead essentially a comment that was way too long for the comment space.

The OP mentioned that he doesn't have a good sense for the shapes of Følner sets in amenable groups of exponential growth. It seems that it might be helpful to give an example.

The simplest amenable group with exponential growth that I am aware of is the Baumslag-Solitar group
$$
BS(1,2) \;=\; \langle a,b \mid bab^{-1} = a^2\rangle
$$
The Cayley graph of this group is made of "sheets" of the following form:

Here red edges correspond to the generator $a$, and blue edges correspond to the generator $b$. These fit together into a binary tree of (slightly offset) sheets:

Note that each sheet is the 1-skeleton of a regular tiling of the hyperbolic plane. (This is the "sixth model" described Cannon, Floyd, Kenyon, and Parry in this paper). It follows that $BS(1,2)$ has exponential growth.

By the way, if you look at the whole Cayley graph from the "front", all of the sheets overlap, but no two vertices lie in the same position on the hyperbolic plane. What this means is that the elements of $BS(1,2)$ are in one-to-one correspondence with *all* the points $(x,y)$ in the hyperbolic plane for which $x$ is a dyadic fraction and $y$ is a power of $2$. Horizontal edges go from $(x,y)$ to $(x+y,y)$, and vertical edges go from $(x,y)$ to $(x,2y)$.

Now, even though $BS(1,2)$ has exponential growth, it turns out to be a solvable group. To see this, note first that $a$ has a sequence of roots
$$
a^{1/2} \;=\; b^{-1}ab,\qquad a^{1/4} \;=\; b^{-2}ab^2,\qquad \ldots
$$
The subgroup of $BS(1,2)$ generated by these roots is normal and is isomorphic to $\mathbb{Z}\bigl[\tfrac{1}{2}\bigr]$, the additive group of the dyadic fractions. Therefore
$$
BS(1,2) \;\cong\; \mathbb{Z}\bigl[\tfrac{1}{2}\bigr] \rtimes \mathbb{Z}
$$
where the $\mathbb{Z}$ is the cyclic subgroup generated by $b$. (This description is of course highly related to the "all in one plane" description of the Cayley graph given above.) This proves that $BS(1,2)$ is solvable, and hence amenable.

Now, the shapes of the balls in $BS(1,2)$ are roughly hyperbolic circles. (More precisely, the intersection of a ball with each sheet is roughly a hyperbolic circle). For example, the following picture shows the intersection of a ball of radius 9 with one sheet of the Cayley graph:

However, the Følner sets in $BS(1,2)$ look nothing like hyperbolic circles. The way to get a Følner set in $BS(1,2)$ is to choose a large "rectangle" in the Cayley graph:
This rectangle must also include the same region in all other "layers" of the Cayley graph that include the bottom edge. More precisely, let
$$
R_{m,n} \;=\; \{(j/2^n,2^k) \mid -m\leq j/2^n \leq m\text{ and }-n\leq k\leq n\}
$$
where we are referring to points in $BS(1,2)$ by their position in the hyperbolic plane. (So $R_{m,n}$ represents a rectangle of "width" $2m$ and "height" $2n$.) Then a sequence of such rectangles will be a Følner sequence as long as the width grows quickly relative to the height. In particular, it is easy to show that
$$
|R_{m,n}| \;=\; (2^{n+1}m+1)(2n+1)
$$
and
$$
|\partial R_{m,n}| \;=\; 2(2^{n+1}m+1)+2(2^{2n+1}-1)
$$
(where $\partial R_{m,n}$ denotes the set of boundary edges), so
$$
\frac{|\partial R_{m,n}|}{|R_{m,n}|} \;=\; \frac{2}{2n+1} + \frac{2(2^{2n+1}-1)}{(2^{n+1}m+1)(2n+1)} \;\sim\; \frac{2}{2n+1} + \frac{2^{n+1}}{m(2n+1)}
$$
where the first term comes from the vertical edges and the second comes from the horizontal edges. As you can see, this will be a Følner sequence as long as $m$ grows at least as quickly as $2^n$.

So the conclusion is that Følner sets in this Cayley graph are very wide rectangles, while the balls are roughly circle-shaped. I am skeptical that there would be any generating set for $BS(1,2)$ that has a subsequence of balls as a Følner sequence. Indeed, no matter what generating set you choose, large enough balls should still be roughly circle-shaped, which means they won't be "wide" enough to be Følner sets.

subsequence$\endgroup$ – Jim Belk Nov 15 '13 at 18:29noamenable groups with exponential growth having a Følner sequence of balls. $\endgroup$ – Jim Belk Nov 15 '13 at 18:472more comments