Is there a workaround for this integral?

One workaround is to shift the main computation from integration to summation. Start from Taylor series for a and get a general SeriesCoefficient, which is much simpler and integrates nicely:

Integrate[SeriesCoefficient[Exp[a Exp[I x]], {a, 0, n}], {x, -Pi, Pi}]

We switched the order of summation and integration. Now what is left is to sum back:

Sum[(2 Sin[n π])/(n n!) a^n,{n,0,∞}]

Out[]= 2 π

This is also quite clear from the fact that for all n>=0 all terms are zero. And the only contributing n=0 term is equal to:

Limit[(2 Sin[n π])/(n n!) a^n,n->0]

Out[]= 2 π

One can use the fact that:

D[Exp[a Exp[I x]], x ] === I a D[Exp[a Exp[I x]], a ]
(* True *)

to integrate by parts and get zero: $$int(a) \equiv \int_{-\pi}^{\pi} dx\ e^{a\ e^{i\ x}}$$ \begin{eqnarray*} \frac{d}{d a} int(a) & = & \int_{-\pi}^{\pi} dx\ e^{i\ x}\ e^{a\ e^{i\ x}} \\ & = & \frac{1}{i a} \int_{-\pi}^{\pi} dx\ \frac{d}{dx} e^{a\ e^{i\ x}} \\ & = & \frac{1}{i a}\ \left[e^{a\ e^{i\ x}} \right]_{-\pi}^{\pi} \\ & = & 0 \end{eqnarray*}

Therefore, at least to first order in a, the integral is independent of a and equal to the value at a=0, i.e. 2 Pi.

The issue here can be identified by evaluating the indefinite integral,

s = Integrate[Exp[a*Exp[I*x]], x]
(* -I ExpIntegralEi[a E^(I x)] *)

Now,

(s /. x -> Pi) - (s /. x -> -Pi)
(* 0 *)

The catch is that ExpIntegralEi[z] has a branch cut, extending along the negative real axis from zero to negative infinity. So, the correct way to apply the limits, at least for a > 0, is

Limit[s, x -> Pi, Direction -> "FromBelow", Assumptions -> a > 0] - 
    Limit[s, x -> -Pi, Direction -> "FromAbove", Assumptions -> a > 0]
(* 2 π *)

as desired. I presume, but have not tried to prove, that employing the proper Limit options will yield the desired result for all a. Certainly,

Integrate[Exp[(-141/10 + 397/100 I) Exp[I*x]], {x, -Pi, Pi}]

and every other specific value of a I have tried yields 2 π.