= Solution
Take $\Gamma=\operatorname{PSL}_2(\mathbb Z)$ as a subgroup of $\operatorname{PSL}_2(\mathbb C)$. It is discrete because representatives have integer entries, so a bounded set of matrices contains only finitely many such representatives, modulo the two choices of sign. It is finitely generated by
$$
S(z)=-1/z,\qquad T(z)=z+1.
$$
The <Euclidean algorithm> on the first column of an integer determinant-one matrix reduces it using these two transformations, proving the generation statement.
It is non-elementary. The powers of $T$ move an interior upper-half-space point towards $\infty$, so $\infty\in\Lambda(\Gamma)$. Conjugating $T$ by $S$ gives a parabolic with fixed point $0$, and conjugating that parabolic by $T$ gives one with fixed point $1$. Their interior orbits approach those respective points. Thus $0,1,\infty$ are three distinct points in the <Kleinian limit set>.
However $S^2=1$ in the projective group, and $S$ is a nonidentity <elliptic Möbius transformation> fixing the hyperbolic geodesic with boundary endpoints $i,-i$. It is a half-turn around that axis. Near an ordinary point of the axis, quotienting identifies the transverse angle modulo $\pi$, giving cone angle $\pi$ instead of $2\pi$. The <elliptic stabilizer obstruction to a hyperbolic manifold quotient> therefore gives
$$
\boxed{\mathbb H^3/\operatorname{PSL}_2(\mathbb Z)
\text{ is a hyperbolic orbifold, not a smooth hyperbolic 3-manifold with its quotient metric}.}
$$
The obstruction is the nonfree action, not finite generation or elementary behavior. An underlying topological quotient can still be a manifold; the conical quotient metric is what prevents this example from being a hyperbolic manifold quotient.
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