= Solution
Work initially with $\operatorname{Re}s>1$, where the <nonholomorphic Eisenstein series> and its differentiated series converge locally absolutely. For $\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in SL_2(\mathbb Z)$, the identities
$$
\operatorname{Im}(\gamma\tau)=\frac y{|c\tau+d|^2},\qquad
m\gamma\tau+n=\frac{(ma+nc)\tau+(mb+nd)}{c\tau+d}
$$
show that each summand at $\gamma\tau$ becomes the summand indexed by $(ma+nc,mb+nd)$ at $\tau$. Reindexing the nonzero integer pairs therefore proves
$$
\boxed{G(\gamma\tau,s)=G(\tau,s).}
$$
Here $y^s$ and the denominator powers use logarithms of positive real numbers, so there is no ambiguity for complex $s$.
For the nonnegative <Laplace-Beltrami operator> of the <hyperbolic plane>, $\Delta y^s=-y^2\partial_y^2y^s=s(1-s)y^s$. If $(m,n)=r(c,d)$ with $(c,d)$ primitive and $r>0$, choose a matrix in $SL_2(\mathbb Z)$ with bottom row $(c,d)$. The corresponding summand is
$$
\frac{y^s}{|m\tau+n|^{2s}}=r^{-2s}\big(\operatorname{Im}(\gamma\tau)\big)^s.
$$
The <Laplace-Beltrami operator> commutes with hyperbolic <isometries>, so this summand has <eigenvalue> $s(1-s)$. This also covers $m=0$, or follows there directly. Termwise differentiation now gives
$$
\boxed{\Delta G(\tau,s)=s(1-s)G(\tau,s).}
$$
The equation describes an automorphic eigenfunction; it does not assert that this noncuspidal function belongs to the square-integrable discrete <spectrum>.
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