Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x  \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$ 

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\frac{K(-\beta,\alpha+1,x)}{\binom{\alpha+\beta}{\beta}}$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes [The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions][1].

See also Rodriguez-like formula on pg. 59 of Bateman's (et al.) [Higher Transcendental Functions Vol. I][2].


  [1]: http://tcjpn.wordpress.com/2011/11/16/a-generalized-dobinski-relation-and-the-confluent-hypergeometric-fcts/
  [2]: http://apps.nrbook.com/bateman/Vol1.pdf