With Maple help I solve the problem:

HoldForm[Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
    m m! (-m + n)!), {n, 0, Infinity}, {m, 1, 
     Infinity}] == -Exp[-l*x]*(-1 + l*(-1 + x))*l*x*
    HypergeometricPFQ[{1, 1, -l*x + l + 2}, {2, 2, -l*x + l + 1}, 
     l*x]] // TeXForm

$$\sum _{n=0}^{\infty } \sum _{m=1}^{\infty } \frac{e^{-l} l^n n (1-x)^{-m+n} x^m}{m m! (-m+n)!}=-\exp (-l x) (-1+l (-1+x)) l x \, _3F_3(1,1,-l x+l+2;2,2,-l x+l+1;l x)$$

For: $\{-1<x<1,l\in \mathbb{R}\}$

ff[x_, l_, M_] := 
 Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
   m m! (-m + n)!), {n, 0, M}, {m, 1, M}] // N

f[x_, l_, M_] := Sum[-((E^-l l^n n (1 - x)^n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
                        x/(-1 + x)])/((-1 + x) (-1 + n)!)), {n, 0, M}] // N;

g[x_, l_] := -Exp[-l*x]*(-1 + l*(-1 + x))*l*x*HypergeometricPFQ[{1, 1, -l*x + l + 2}, {2, 2, -l*x + l + 1}, l*x]

ff[1/2, -1, 20]
(*-0.282835*)
f[1/2, -1, 20]
(*-0.282835*)
g[1/2, -1] // N
(*-0.282835*)

Do you have any reason to think there is a closed form?

Most Sum don't have one. Maybe the best you can do is numerical methods.

Maybe You put the question here more likely you will get someone to help you.

  Sum[(E^-l l^n n (1 - x)^(-m + n) x^m)/(
   m m! (-m + n)!), {m, 1, Infinity}, 
   Assumptions -> {0 < x < 1, 0 < l < 1, n >= 0}]
  (* -((E^-l l^n n (1 - x)^
    n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
     x/(-1 + x)])/((-1 + x) (-1 + n)!))*)

 Sum[-((E^-l l^n n (1 - x)^
    n x HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, 
     x/(-1 + x)])/((-1 + x) (-1 + n)!)), {n, 0, Infinity}, 
  Assumptions -> {0 < x < 1, 0 < l < 1}] (*Can't Find !*)

Regards M.I.