= Solution
For $m=0$ the two normalizations agree, and the <Laguerre polynomials> are orthonormal with weight $e^{-x}$. Take real $-1<h<1$. Their expansion coefficients for the requested function are
$$
c_n(h)=\frac1{1-h}\int_0^\infty e^{-x/(1-h)}L_n(x)\,dx.
$$
Substitute <Rodrigues' formula> and integrate by parts $n$ times. Differentiating $e^{-xh/(1-h)}$ transfers a factor $[h/(1-h)]^n$ after the two sign factors cancel. Therefore
$$
c_n(h)=\frac1{(1-h)n!}\left(\frac h{1-h}\right)^n\int_0^\infty x^ne^{-x/(1-h)}\,dx
=\boxed{h^n}.
$$
The integral is $n!(1-h)^{n+1}$ by the <Gamma function>. Boundary terms vanish throughout this range.
The resulting <generating function for Laguerre polynomials> is
$$
\boxed{\frac{e^{-xh/(1-h)}}{1-h}=\sum_{n=0}^\infty h^nL_n(x),\qquad |h|<1.}
$$
One can also verify convergence and equality directly: <Rodrigues' formula> gives $L_n(x)=\sum_{j=0}^n\binom nj(-x)^j/j!$, and summing $\sum_{n\geq j}\binom njh^n=h^j/(1-h)^{j+1}$ reduces the double series to the exponential on the left. The absolute-value double series converges for fixed $x$ and $|h|<1$, justifying the interchange; the formula then also holds for complex $h$ in that disk.
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