note that $$e^{a\partial/\partial x}f(x)=f(x+a)$$ is the translation operator, so your exponent of the delta function gives $2\pi \delta(x-in)$, which is indeed consistent with
$$\int_{-\infty}^\infty e^{izy}dy=2\pi\delta(z)$$
for $z=x-in$.
All of this is purely formal, but there have been attempts to put it on a more secure ground by defining the delta function of a complex argument as an "ultradistribution". See for example Theories of Generalised Functions (page 121) and Distribution Theory and Transform Analysis (page 204). Caveat emptor.