It is very easy to prove that if $D=\dfrac{d}{dx}$, then $(e^{nD}f)(x)=f(x+n)$ about $x=m$ in the real numbers.
Proof: $$(e^{mD}f)=\sum^\infty_{n=0}\dfrac{D^nf}{n!}m^n\\ \implies (e^{mD}f)(x)=\sum^\infty_{n=0}\dfrac{f^{(n)}(x)}{n!}m^n$$ This is the Taylor expansion $f(x+m)$ about $x=m$.
Let $D=\dfrac{\partial}{\partial x}$. Then, consider $(e^{-iD}f)(x)$. Is there an analog of the above equation for imaginary operators? If no, why? (My confusion lies in the fact that $f$ must have a real input and output.)
Yes, there is an analogue. Just as the operator $e^{aD}$ performs a spatial translation on the position where the function is being evaluated, the operator $e^{iaD}f)$ performs a translation in momentum space on the momentum at which the function is being evaluated:
$$(e^{iaD}f)(x)=f(p+a).$$
That is to say, it shifts the Fourier transform of a function by some amount $a$ instead of shifting the function itself.