= Solution
For a finite symmetric generating alphabet $\Sigma$, the <growth function of a finitely generated group> is
$$
\boxed{\beta_{G,\Sigma}(n)=|B_\Sigma(1,n)|=\#\{g\in G:|g|_\Sigma\le n\}}.
$$
It counts distinct elements, not words. Relations affect which words coincide, while the resulting coarse growth type is independent of the finite generating set.
Let $f:G\to H$ be a <quasi-isometry> between the vertex <word metrics>, with constants $\lambda,\epsilon$. A <quasi-isometry> of the metric graphs can be rounded to vertices with only bounded changes of these constants. If $f(g)=f(g')$, its lower distortion bound gives $d_G(g,g')\le\lambda\epsilon$. Thus every nonempty fibre has at most $M=\beta_G(\lceil\lambda\epsilon\rceil)$ elements, independently of its centre. Also $f(B_G(1,n))$ lies in the ball of radius $\lambda n+\epsilon$ about $f(1)$. Left translation in $H$ preserves ball cardinalities, so
$$
\beta_G(n)\le M\beta_H(\lceil\lambda n+\epsilon\rceil).
$$
Apply the same argument to a <quasi-inverse of a quasi-isometry> to get the reverse comparison. Enlarging all constants to an integer $K$ gives the two inequalities required by the paper's $\lesssim$ relation, and therefore \b[$\beta_G\sim_e\beta_H$]. This proves <quasi-isometry> invariance of the stated growth class.
For the exponential upper bound, put $q=|\Sigma|$. Every element in the radius-$n$ ball has a representative of length at most $n$. Hence
$$
\beta_G(n)\le\sum_{j=0}^nq^j\le(q+1)^n\qquad(n\ge0),
$$
where the last inequality follows by comparison with the binomial expansion of $(q+1)^n$; it also holds for the empty alphabet of the trivial group. Thus \b[every <finitely generated group> has at most exponential growth], with no finite-relator assumption needed.
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