Looks like :

Integrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*Exp[Exp[t]], {t, -Pi, Pi}] ==
((-1)^(I (n + \[Lambda])) (-Gamma[-I (n + \[Lambda]), -E^\[Pi]] + 
Gamma[-I (n + \[Lambda]), -Cosh[\[Pi]] + Sinh[\[Pi]]]))/(2 \[Pi])

We can check:

f[n_, \[Lambda]_] := 
 NIntegrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*
 Exp[Exp[t]], {t, -Pi, Pi}, Method -> "LocalAdaptive", 
 WorkingPrecision -> 25];
 g[n_, \[Lambda]_] := ((-1)^(
 I (n + \[Lambda])) (-Gamma[-I (n + \[Lambda]), -E^\[Pi]] + 
 Gamma[-I (n + \[Lambda]), -Cosh[\[Pi]] + Sinh[\[Pi]]]))/(
 2 \[Pi]); N[g[1, 1] - f[1, 1], 20]

  (*0.*10^-18 + 0.*10^-19 I*) (* Looks good *)

Mathematica is not a magic box that'll spit out a solution to any problem.

All computer algebra systems, including Mathematica, are limited in their capabilities.

This is the integral to evaluate

(1/(2*Pi))*
 Integrate[Exp[-(I*(n + \[Lambda])*t)]*Exp[Exp[t]], {t, -Pi, Pi}]

Yes, I expanded the exponential and summarized the terms as you did.

Sum[Integrate[(1/(2*Pi))*Exp[-(I*(n + \[Lambda])*t)]*(Exp[t])^j/
   j!, {t, -Pi, Pi}], {j, 0, Infinity}]

(*-(((-1)^(I n + 
   I \[Lambda]) (Gamma[-I n - I \[Lambda], 0, -E^-\[Pi]] - 
    Gamma[-I (n + \[Lambda]), 0, -E^\[Pi]]))/(2 \[Pi]))*)