= Solution
Keep $b=t^{-1}at$. Repeatedly applying $bab^{-1}=a^2$ gives
$$
b^na b^{-n}=a^{2^n}.
$$
In the <word metric> on $G$ from the finite generating set $\{a,t\}$, the right-hand side therefore has length at most $6n+1$, because $b$ is represented by the word $t^{-1}at$. In the intrinsic word metric on the <infinite cyclic group> $\langle a\rangle$ generated by $a$, however,
$$
|a^{2^n}|_{\langle a\rangle}=2^n.
$$
A <quasi-isometric embedding> would bound the latter by an affine function of the former. The exponential sequence above violates every such bound, so the inclusion $\langle a\rangle\hookrightarrow G$ is not a quasi-isometric embedding.
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