Growth type is a quasi-isometry invariant of finitely generated groups

By composing $f$ with left translation by $f(e{)}^{-1}$ in $H$, which is an isometry by [L4], we may assume $f(e)=e$ without changing any fiber cardinalities or quasi-isometry constants up to harmless enlargement.

By [L1], choose a coarse Lipschitz quasi-inverse $q:H\to G$ and a real $c\ge 0$ with $d_S(q(f(g)),g)\le c$ for all $g\in G$. By [L2], enlarge constants so that both $f$ and $q$ are Lipschitz, say $d_T(f(g),f(h))\le L{d}_S(g,h)$ and $d_S(q(u),q(v))\le L^{\mathrm{\prime }}{d}_T(u,v)$.

Let $A:=\lceil L\rceil$, $B:=\lceil L^{\mathrm{\prime }}\rceil$, $m_G:={\beta }_{G,S}(\lceil 2c\rceil )$, and $m_H:={\beta }_{H,T}(\lceil 2c\rceil )$. Then $A,B,m_G,m_H$ are natural numbers, and the Lipschitz bound on $f$ gives $f(B_S(e,n))\subseteq B_T(e,An)$ for every $n$. If $f(g_1)=f(g_2)$, then step 1.2 gives $d_S(g_1,g_2)\le 2c\le \lceil 2c\rceil$, so every fiber of $f$ has size at most $m_G$, finite by [L3]. Therefore ${\beta }_{G,S}(n)\le m_G{\beta }_{H,T}(An)$.

Applying the same argument to the quasi-inverse $q$ gives ${\beta }_{H,T}(n)\le m_H{\beta }_{G,S}(Bn)$ for all $n$.

Let $K$ be a natural number with $K\ge A,B,m_G,m_H$. Growth functions are nondecreasing and every radius-$n$ word-metric ball contains the identity, so step 2.1 gives ${\beta }_{G,S}(n)\le m_G{\beta }_{H,T}(An)\le K{\beta }_{H,T}(Kn+K)+K,$ and step 2.2 similarly gives ${\beta }_{H,T}(n)\le K{\beta }_{G,S}(Kn+K)+K\mathrm{.}$ These are the two comparison directions of [L5]. Hence ${\beta }_{G,S}\simeq {\beta }_{H,T}$, so $G$ and $H$ have the same growth type.