= Solution
Use the <primitive real-analytic Eisenstein series> with exactly the normalization in the PDF. Its <coset> expression uses $\Gamma_\infty=\{\pm\begin{pmatrix}1&n\\0&1\end{pmatrix}:n\in\mathbb Z\}$, so primitive bottom rows differing by sign represent the same <coset>. The factor $1/2$ in the primitive-pair sum accounts for this sign identification. Unlike the sum over all nonzero <integer> pairs, this <Eisenstein series> has no additional factor $2\zeta(2w)$. Write $w$ for the <Rankin–Selberg integral> parameter and $s$ for the <Dirichlet series> parameter to keep their shift visible.
First establish an initial region where all rearrangements are absolute. The <invariant norm of a modular form> $y^{k/2}|f(\tau)|$ is bounded: it is $\Gamma$-invariant, continuous on a truncated fundamental domain, and tends to <zero> at the cusp. Integrating one period of the <Fourier series> gives
$$
a_ne^{-2\pi ny}=\int_0^1 f(x+iy)e^{-2\pi inx}\,dx,
\qquad |a_n|\leq Cy^{-k/2}e^{2\pi ny}.
$$
With $y=1/n$, this proves the <Fourier coefficient bound for a cusp form> $a_n=O(n^{k/2})$, and similarly $b_n=O(n^{k/2})$. Hence $D(f,g,s)$ is initially absolutely convergent for $\operatorname{Re}s>k+1$.
For $\operatorname{Re}w>1$, the <primitive real-analytic Eisenstein series> is locally uniformly convergent and <holomorphic> in $w$. On the standard fundamental domain it has at most <polynomial growth> in $y$, locally uniformly in this half-plane. These are the analytic properties of the <Eisenstein series> needed here; they can also be checked directly from its defining sum. For example, for real $\sigma>1$ and $y\geq1$, majorize the primitive sum by the sum over all nonzero pairs and use
$$
\sum_{n\in\mathbb Z}\big((mx+n)^2+(my)^2\big)^{-\sigma}
\leq C_\sigma\big((|m|y)^{-2\sigma}+(|m|y)^{1-2\sigma}\big)
\quad(m\ne0).
$$
This bound follows by comparing a shifted one-dimensional sum with its integral and a largest summand. Summing over $m\ne0$ gives $E(\tau,\sigma)=O(y^\sigma+y^{1-\sigma})$. For complex $w$, $|E(\tau,w)|\leq E(\tau,\operatorname{Re}w)$, and the estimates are uniform on compact parameter sets. Since $f$ and $g$ decay exponentially on the cusp end, the integral $I(f,g,w)$ is therefore <holomorphic> throughout $\operatorname{Re}w>1$ by dominated integration.
The product $f(\tau)\overline{g(\tau)}y^k$ is $\Gamma$-invariant: the two modular factors cancel the transformation $\operatorname{Im}(\gamma\tau)=y/|c\tau+d|^2$. With the <hyperbolic area> $d\mu=dx\,dy/y^2$, <Rankin–Selberg unfolding> gives, initially for $\operatorname{Re}w>2$,
$$
\begin{aligned}
I(f,g,w)
&=\int_{\Gamma\backslash\mathbb H}f\overline g\,y^k
\sum_{\gamma\in\Gamma_\infty\backslash\Gamma}(\operatorname{Im}\gamma\tau)^w\,d\mu\\
&=\int_{\Gamma_\infty\backslash\mathbb H}f(\tau)\overline{g(\tau)}y^{w+k}\,d\mu\\
&=\int_0^\infty\int_0^1 f(x+iy)\overline{g(x+iy)}y^{w+k-2}\,dx\,dy.
\end{aligned}
$$
Absolute unfolding is justified by replacing $f\overline g$ by $|fg|$ and using the same finite fundamental-domain integral with $E(\tau,\operatorname{Re}w)$. The subsequent termwise integrations can be justified particularly economically by <Cauchy-Schwarz inequality>: on each horizontal line,
$$
\int_0^1|f(x+iy)g(x+iy)|\,dx
\leq\left(\sum_{n\geq1}|a_n|^2e^{-4\pi ny}\right)^{1/2}
\left(\sum_{n\geq1}|b_n|^2e^{-4\pi ny}\right)^{1/2}.
$$
The coefficient bounds make the right side $O(y^{-k-1})$ as $y\downarrow0$, and it decays exponentially as $y\to\infty$. Multiplication by $y^{\operatorname{Re}w+k-2}$ is integrable at <zero> when $\operatorname{Re}w>2$. Thus the integral of the inner <Fourier series> may be evaluated first. <Orthogonality> of the exponentials gives
$$
\int_0^1 f(x+iy)\overline{g(x+iy)}\,dx
=\sum_{n\geq1}a_n\overline{b_n}e^{-4\pi ny}.
$$
The <Mellin transform> of each exponential is
$$
\int_0^\infty e^{-4\pi ny}y^{w+k-2}\,dy
=\frac{\Gamma(w+k-1)}{(4\pi n)^{w+k-1}}.
$$
In the same initial region the sum of the absolute values of these integrals is finite by $a_n\overline{b_n}=O(n^k)$. Consequently the exact <Rankin–Selberg integral> identity is
$$
\boxed{I(f,g,w)=\frac{\Gamma(w+k-1)}{(4\pi)^{w+k-1}}D(f,g,w+k-1).}
$$
In particular the two parameters are not equal: the <Dirichlet series> argument is shifted by $k-1$.
The holomorphy of $I$ on $\operatorname{Re}w>1$ now gives the required continuation:
$$
\boxed{D(f,g,s)=\frac{(4\pi)^s}{\Gamma(s)}I(f,g,s-k+1),\qquad \operatorname{Re}s>k.}
$$
The right side agrees with the original <Dirichlet series> on $\operatorname{Re}s>k+1$ and is <holomorphic> on the larger half-plane. No continuation of the <Eisenstein series> beyond $\operatorname{Re}w>1$ is needed.
It is important to supply a further argument for <absolute convergence>: analytic continuation alone does not imply it. For real $w>1$, both $E(\tau,w)$ and $|f(\tau)|^2$ are nonnegative, and their fundamental-domain integral is finite by the cusp estimates above. <Tonelli theorem> therefore unfolds $I(f,f,w)$ even before knowing convergence of the <Dirichlet series>. On each fixed horizontal line, the <Fourier series> converges absolutely and uniformly, so integration in $x$ gives
$$
\int_0^1|f(x+iy)|^2\,dx=\sum_{n\geq1}|a_n|^2e^{-4\pi ny}.
$$
A second use of <Tonelli theorem>, now in $y$, proves the <positive unfolding of a cusp-form square>:
$$
I(f,f,w)=\frac{\Gamma(w+k-1)}{(4\pi)^{w+k-1}}
\sum_{n\geq1}\frac{|a_n|^2}{n^{w+k-1}}<\infty.
$$
Since every real $s>k$ has $s=w+k-1$ with $w>1$, we have \b[<absolute convergence> of $D(f,f,s)$ for $\operatorname{Re}s>k$]. <Absolute convergence> for complex $s$ follows by taking its real part. In fact <Cauchy-Schwarz inequality> also gives <absolute convergence> of $D(f,g,s)$ there from the two corresponding square sums.
Finally let $\sigma>(k+1)/2$ and choose $k<\beta<2\sigma-1$. The preceding square-coefficient bound and <Cauchy-Schwarz inequality> give
$$
\sum_{n\geq1}|a_n|n^{-\sigma}
\leq\left(\sum_{n\geq1}|a_n|^2n^{-\beta}\right)^{1/2}
\left(\sum_{n\geq1}n^{-(2\sigma-\beta)}\right)^{1/2}<\infty.
$$
This proves <absolute convergence of cusp-form L-series from square coefficients>, and hence
$$
\boxed{L(f,s)\text{ converges absolutely for }\operatorname{Re}s>\frac{k+1}{2}.}
$$
Equivalently, the suggested elementary inequality works by choosing $k-\sigma<\alpha<\sigma-1$: after multiplication by $n^{-\sigma}$ its two bounds sum to a convergent $\sum n^{-(\sigma-\alpha)}$ and a convergent $\sum|a_n|^2n^{-(\sigma+\alpha)}$. The existence of this interval is precisely $2\sigma>k+1$.
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