= Solution
A map $f:X\to Y$ is a <quasi-isometry> if there are $\lambda\geq1$ and $\varepsilon,R\geq0$ such that
$$
\lambda^{-1}d_X(x,x')-\varepsilon\leq d_Y(fx,fx')\leq\lambda d_X(x,x')+\varepsilon
$$
for all $x,x'\in X$, and every point of $Y$ lies within distance $R$ of $f(X)$. The <Milnor–Švarc lemma> says that a group acting properly discontinuously, cocompactly and isometrically on a proper geodesic metric space is finitely generated, and each orbit map from a <word metric> is a quasi-isometry.
Now let $S,T$ be finite generating sets of $G$. Put
$$
L=\max_{s\in S}|s|_T,
\qquad
M=\max_{t\in T}|t|_S.
$$
Replacing each letter in an $S$-word by a $T$-word and conversely gives
$$
M^{-1}d_S(g,h)\leq d_T(g,h)\leq Ld_S(g,h).
$$
Thus the identity map is a <bilipschitz equivalence>, hence a quasi-isometry. \b[All finite generating sets of a finitely generated group give quasi-isometric Cayley graphs.]
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