Completeness of an orthogonal sequence of functions is a bit tricky on unbounded intervals, while it is relatively straightforward on bounded intervals. In the case of Laguerre and Hermite polynomials, there is a nice trick due to von Neumann that allows the reduction to bounded intervals.
There seems to be a bit of confusion about the interval in the statement of the question. Here's a correct statement:
For any real number $\alpha \gt -1$ the functions $\langle e^{-x/2} x^{\alpha/2} L_{n}^{(\alpha)}(x)\rangle_{n=0}^\infty$ obtained from the Laguerre polynomials $L_{n}^{(\alpha)}(x)$ are a complete orthogonal system in $L^2(0,\infty)$. The Hermite polynomials $H_n(x)$ yield the complete orthogonal system $\langle e^{-x^2/2} H_n(x)\rangle_{n=0}^\infty$ in $L^2(\mathbb{R})$.
This is proved in detail in the classic book Gábor Szegő, Orthogonal polynomials, Chapter 5. The entire chapter discusses the main properties oft the Laguerre polynomials $L^{(\alpha)}_n(x)$ for an arbitrary real number $\alpha \gt -1$ and proves their completeness in Section 5.7.
More precisely, Szegő shows in Theorem 5.7.1 on pages 108f that for fixed $\alpha \gt -1$ the functions $f_n(x) = e^{-x/2}x^{\alpha/2} x^n$ span a dense subspace of $L^2(0,\infty)$.
The first idea is to use a change of variables $y = e^{-x}$ in order to use the case of $L^2(0,1)$ where density of the span of $(\log1/y)^{\alpha/2} y^n$ is not too hard to prove (see Theorem 3.1.5).
Write a function in $L^2(0,\infty)$ as $e^{-x/2} x^{\alpha/2} f(x)$. Then we have that $(\log1/y)^{\alpha/2} f(\log(1/y)) \in L^2(0,1)$ can be approximated by functions of the form $(\log1/y)^{\alpha/2} p(y)$ where $p$ is a polynomial. Transforming back to $(0,\infty)$ this shows that
$$
\int_{0}^\infty e^{-x} x^\alpha (f(x) - p(e^{-x}))^2 \,dx \lt \varepsilon
$$
for a suitable polynomial $p$. This reduces the task to proving that for all natural $k$ there exists a polynomial $q$ such that
$$\tag{$\ast$}
\int_{0}^\infty e^{-x} x^\alpha (e^{-kx} - q(x))^2\,dx
$$
is as small as we wish.
To do this, von Neumann's trick is to use the generating function
of the Laguerre polynomials $L_{n}^{(\alpha)}(x)$
$$
(1-w)^{-\alpha-1} \exp\left(-\frac{xw}{1-w}\right) = \sum_{n=0}^\infty L_n^{(\alpha)}(x) w^n.
$$
Choosing $w = \frac{k}{k+1}$ we have $\exp\left(-\frac{xw}{1-w}\right) = \exp{(-kx)}$.
Thus, a natural choice for $q$ is $q_N(x) = (1-w)^{\alpha+1} \sum_{n=0}^N L_n^{(\alpha)}(x) w^n$ with large enough $N$. Plugging this into $(\ast)$ we obtain using the orthogonality relations
$$
\begin{align*}
\int_{0}^\infty e^{-x} x^\alpha (e^{-kx} - q_N(x))^2\,dx & = (1-w)^{2\alpha+2} \int_{0}^\infty e^{-x} x^\alpha \left(\sum_{n=N+1}^\infty L_{n}^{(\alpha)}(x) w^{n}\right)^2\,dx \\
&= (1-w)^{2\alpha+2} \Gamma(\alpha+1) \sum_{n=N+1}^\infty \binom{n+\alpha}{n} w^{2n}
\end{align*}$$
where term-wise integration is justified using an application of Cauchy-Schwarz. It remains to observe that the last expression tends to $0$ as $N \to \infty$.
Another reference discussing the case of $\alpha = 0$ nicely is Courant and Hilbert, Methods of mathematical physics, I, §9, sections 5 and 6. They discuss ordinary Laguerre and Hermite polynomials and their completeness.
Finally I found how to do it. I post it, if someone is interested.
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}-\varkappa^2/4}e^{\varkappa^2/4}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4}\underbrace{\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-(\tilde{x}-\varkappa/2)^2}H_l(\tilde{x})\;\mathrm{d}\tilde{x}}_I
\end{align}
If we pose $x = \tilde{x}-\frac{\varkappa}{2}$ in this expression, the integral $I$ becomes
\begin{equation*}
I = \int_{-\infty}^{+\infty}H_n(x+\varkappa/2)e^{-x^2}H_l(x+\varkappa/2)\;\mathrm{d}x
\end{equation*}
We know that
\begin{equation*}
H_n(x+a) = \sum_{p=0}^n \frac{n!}{(n-p)!p!}(2a)^{n-p}H_p(x)
\end{equation*}
Hence, the integral $I$ becomes
\begin{align*}
I &= \int_{-\infty}^{+\infty} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\varkappa^{n-p}H_p(x) e^{-x^2} \sum_{q=0}^l \frac{l!}{(l-q)!q!}\varkappa^{l-q}H_q(x)\;\mathrm{d}x \\
&= \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\varkappa^{n-p}\frac{l!}{(l-q)!q!}\varkappa^{l-q}\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x \\
\end{align*}
The Hermite polynomials are orthogonal in the range $(-\infty,\infty)$ with respect to the weighting function $e^{-x^2}$ and satisfy
\begin{alignat*}{2}
&&&\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x = \sqrt{\pi}2^pp!\;\delta_{pq} \\
&\Rightarrow\quad&& I = \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\frac{l!}{(l-q)!q!}\varkappa^{n+l-p-q}\cdot \sqrt{\pi}2^pp!\;\delta_{pq}
\end{alignat*}
As this integral is nil if we have not $p=q$, we can replace the two sums by only one that goes from 0 to $\min(n,l)$. Let us say that $n<l$. Hence, the full expression for the $D$-matrix is
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!p!}2^pp!\sqrt{\pi}\;\varkappa^{n+l-2p} \notag\\
&= \frac{\varkappa^{n+l}}{\sqrt{2^nn!}\sqrt{2^ll!}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!}2^p\;\varkappa^{-2p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{n+l}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{-p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{n-p}
\end{align}
Associated Laguerre polynomials $L_a^b(x)$ are given by
\begin{equation*}
L_a^b(x) = \sum_{k=0}^{a}(-1)^k \frac{(a+b)!}{(a-k)!(b+k)!k!}x^k
\end{equation*}
It suggests us to transform the expression of the $D$-matrix by posing $k=n-p$. Hence, we have
\begin{align}
D_{ln}(\varkappa) &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=n}^0 \frac{(l)!}{(n-(n-k))!(l-(n-k))!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{n-(n-k)} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n \frac{l!}{k!(l-n+k)!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n (-1)^k\frac{([l-n]+n)!}{(n-k)!([l-n]+k)!k!}\left(-\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4}L_n^{l-n}\left(-\frac{\varkappa^2}{2}\right)
\end{align}
It should be remembered that we had supposed that $n<l$. But that could be otherwise. In order to be general, $n_<$ and $n_>$ will be defined as $n_<=\min{(n,l)}$ and $n_>=\max{(n,l)}$ and $l-n=|l-n|$. We then have
\begin{equation}
D_{ln}(\varkappa) = \sqrt{\frac{n_<!}{n_>!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{|l-n|}L_{n_<}^{|l-n|}\left(-\frac{\varkappa^2}{2}\right)e^{\varkappa^2/4}
\end{equation}
Best Answer
Then those tests are sufficient to show that the result is wrong, and there's nothing else to say there.
You can't. The degree of the polynomial on the left is different to that on the right whenever $n\neq m$.
Then the paper is wrong, at least as regards this particular result. From a cursory reading of the subsequent text, it doesn't seem to be crucial to the rest of the arguments presented, so you should be able to fix any problems with it, if you really need to.