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.
Best Answer
This is wrong (e.g., it's clearly wrong for $m=0$). The $x$ outside the sum should be a $t$:
$$ \frac1{1-t}\left(-\frac t{1-t}\right)^m\mathrm e^{-\frac{xt}{1-t}}=\sum_{n=m}^\infty L_n^m(x)\frac{t^n}{n!}\;, $$
which follows directly by differentiating $f(t,x)$ $m$ times with respect to $x$.