The Grigorchuk group is finitely generated and has subexponential non-polynomial growth but I'm not aware of a finite presentation. Does a finite presentation imply that the group is polynomial or exponential as well?
Is there a known example of a finitely presented group with subexponential growth that isn’t polynomial
geometric-group-theorygroup-presentationgroup-theorysubgroup-growth
Related Solutions
Let $K=\ker \pi$ and, as in your post, let $G = \langle S, g \rangle$, with $S$ a finite subset of $K$. Then $K = \langle g^i S g^{-i} : i \in {\mathbb Z} \rangle$.
For $N \ge 0$, let $K_N = \langle g^i S g^{-i} : 0 \le i \le N \rangle$, and for $N \le 0$ let $K_N = \langle g^{-i} S g^i : 0 \le i \le -N \rangle$.
So we have two ascending chains of subgroups of $K$: $K_0 \le K_1 \le K_2 \le \cdots$ and $K_0 \le K_{-1} \le K_{-2} \le \cdots $.
If both of these chains stabilize after finitely many steps, then $K$ is generated by a finite number of the $K_i$ and hence $K$ is finitely generated, and we are done. So we can assume that one of them does not stabilize, and by swapping $g$ and $g^{-1}$ if necessary, we can assume that $K_0 \le K_1 \le K_2 \le \cdots$ does not stabilize.
Suppose that, for some $s \in S$ and $N_s > 0$, we have $$g^{N_s} s g^{-N_s} \in \langle s,gsg^{-1},\ldots,g^{N_s-1}sg^{-(N_s-1)} \rangle.$$ Then the same condition holds for all $N \ge N_s$.
If such an $N_s$ existed for all $s \in S$, then the condition would hold for all $N \ge \max(N_s)$ and all $s \in S$. But then we would have $K_n \le K_{n-1}$ for all $n \ge \max(N_s)$, and the chain $K_0 \le K_1 \le K_2 \le \cdots$ would stabilize, contrary to assumption.
So there exists $s \in S$, such that $g^n s g^{-n} \notin \langle s, gsg^{-1},\ldots, g^{n-1}sg^{-(n-1)} \rangle$ for all $n > 0$.
We claim that the subsemigroup of $G$ generated by $gs$ and $g^2$ is free, which implies that $G$ has exponential growth.
If not then let $w_1$ and $w_2$ be distinct (positive) words in $g^2$ and $gs$ of smallest total length such that $w_1 =_G w_2$. Then one of $w_1,w_2$ - say $w_1$ - must end in $gs$ and the other in $g^2$, since otherwise we could cancel the final letters and get shorter equal words.
Now, since $\pi(w_1) = \pi(w_2)$, they must both contain the same number of occurrences of $g^2$ and of $gs$ - suppose there is a total of $n$ occurrences of $g$ in both words.
Then, when we rewrite $w_1$ and $w_2$ in $G$ to collect the powers of $g$ to the right (as you have done in your post), we get $w_1' g^{n} = w_2'g^n$, where $w_1'$ and $w_2'$ are words in conjugates $g^ksg^{-k}$ of $s$ for $k \ge 0$.
Since $w_1$ ends in $gs$, and $w_2$ ends in $g^2$, the largest such $k$ occurring in $w_1'$ will be $g^nsg^{-n}$ at the end of $w'$, but the largest $k$ in $w_2$ will be less than $n$. So we get $g^n s g^{-n} \in \langle s, gsg^{-1},\ldots, g^{n-1}sg^{-(n-1)} \rangle$, contrary to assumption.
Best Answer
There are no known examples of such groups. Grigorchuk group is infinitely presented and so are all other known infinite finitely generated groups of intermediate growth (there are many examples: Gupta-Sidki, Erschler, and others).