= Solution
Write $d\mu=dx\,dy/y^2$ and distinguish the all-pairs <Eisenstein series> in this problem from its primitive version
$$
E_0(z,w)=\frac12\sum_{(c,d)=1}\frac{y^w}{|cz+d|^{2w}}
=\sum_{\Gamma_\infty\backslash SL_2(\mathbb Z)}(\operatorname{Im}\gamma z)^w.
$$
Separating each nonzero integer pair into a positive integer multiple of a primitive pair gives
$$
E(z,w)=2\pi^{-w}\Gamma(w)\zeta(2w)E_0(z,w).
$$
The factor two accounts for the two signs of a primitive pair. It must not be omitted in the <Rankin–Selberg unfolding>.
Because $f\overline g y^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\leq x<1$, $y>0$:
$$
\int_{\mathcal F}f(z)\overline{g(z)}y^kE_0(z,w)d\mu
=\int_0^\infty\int_0^1 f(x+iy)\overline{g(x+iy)}y^{w+k-2}dx\,dy.
$$
The $x$ integral of the product <Fourier expansions> is $\sum_{n\geq1}a(n)\overline{b(n)}e^{-4\pi ny}$. The <Gamma function> integral then gives
$$
\int_{\mathcal F}f\overline g y^kE_0(z,w)d\mu
=\frac{\Gamma(w+k-1)}{(4\pi)^{w+k-1}}F(w+k-1).
$$
These steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and <modular cusp> decay show the square-coefficient series converges for $\operatorname{Re}s>k$; the <Cauchy-Schwarz inequality> gives the same absolute-convergence range for the mixed series.
Put $w=s-k+1$ and define
$$
\boxed{\mathcal R(s):=2\pi^{-w}\Gamma(w)\zeta(2w)\frac{\Gamma(s)}{(4\pi)^s}F(s)
=\int_{\mathcal F}f(z)\overline{g(z)}y^kE(z,w)d\mu.}
$$
This is the <zeta-completed Rankin-Selberg coefficient series>. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.
The exponential decay of <cusp forms> at infinity dominates the <polynomial> growth in $y$ of the <Eisenstein series> and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb C$, with possible simple poles only at $s=k-1,k$, and its <functional equation> is
$$
\boxed{\mathcal R(s)=\mathcal R(2k-1-s).}
$$
Indeed $w\mapsto1-w$ corresponds to $s\mapsto2k-1-s$. If $\langle f,g\rangle=0$, both pole residues vanish and the completion is entire.
The normalization of the poles can also be made explicit. The constant term of the given <Eisenstein series> is
$$
2\pi^{-w}\Gamma(w)\zeta(2w)y^w
+2\pi^{1/2-w}\Gamma(w-\tfrac12)\zeta(2w-1)y^{1-w}.
$$
The second term has residue one at $w=1$, since the pole of $\zeta(2w-1)$ has residue $1/2$; the <functional equation> gives residue minus one at $w=0$. Hence the residues of $\mathcal R$ at $k$ and $k-1$ are respectively $\langle f,g\rangle$ and $-\langle f,g\rangle$, the <Petersson inner product>.
It is important to separate this completed statement from the raw $F$. Inverting the completion gives <meromorphic continuation> of $F$, but zeros of $\zeta(2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow2k-1-s$. At the two displayed completion poles, for a nontrivial <modular cusp> <modular weight>,
$$
\boxed{\operatorname*{Res}_{s=k}F(s)=\frac{3(4\pi)^k}{\pi\Gamma(k)}\langle f,g\rangle,
\qquad F(k-1)=\frac{(4\pi)^{k-1}}{\Gamma(k-1)}\langle f,g\rangle.}
$$
The latter is finite because the pole of $\Gamma(w)$ cancels that of the integral at $w=0$, using $\zeta(0)=-1/2$. In particular, taking $f=g\ne0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal R$, not the raw series with all completion factors suppressed.
Finally suppose the two $L$-series have <Euler products>. With first <coefficients> normalized to one, their <coefficients> are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Therefore
$$
\boxed{F(s)=\prod_p\left(\sum_{r\geq0}a(p^r)\overline{b(p^r)}p^{-rs}\right),}
$$
initially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized <coefficients> inside it.
For the usual degree-two Hecke products of Question 3, take roots $\alpha_p,\beta_p$ with sum $a(p)$ and product $p^{k-1}$, and $\gamma_p,\delta_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=(\alpha_p^{r+1}-\beta_p^{r+1})/(\alpha_p-\beta_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the <Euler factor of a coefficientwise product of Hecke eigenforms>:
$$
\boxed{F_p(X)=\frac{1-p^{2k-2}X^2}
{(1-\alpha_p\overline\gamma_pX)(1-\alpha_p\overline\delta_pX)
(1-\beta_p\overline\gamma_pX)(1-\beta_p\overline\delta_pX)},\qquad X=p^{-s}.}
$$
Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta(2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.
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