Exponential of the differential operator

We first make a heuristic calculation.

The equation in question is $Df = e^{D}f$. Thus it suffices to solve that $D = e^D$. Treating $D$ as a complex number, it is not hard to see that this equation has a zero in $\Bbb{C}$. For example, we may exploit Lambert-W function to specify such a number. Now let $r$ satisfy $r = e^r$. Then the function $f(t) = e^{r t}$ satisfies the desired property.

Indeed, we prove that this is the case. We have

$$Df(t) = r e^{r t} = e^{r}e^{r t} = e^{r(t+1)} = f(t+1) = e^{D}f(t).$$

Therefore $f(t)$ satisfies the desired property.

If we confine ourselves to real-valued functions, taking real part or imaginary part will also give such non-trivial examples.

Let us elaborate on the differential operator $ {\rm e}^{aD} $. It is defined by the power series $ \sum_{k=0}^{\infty}\frac{D^k}{k!} $. We would like to know its effect on the function $f(x)$. First let's see its effect on the function $x^m$. $$ {\rm e}^{ aD } x^m = \sum_{k=0}^{\infty} \frac{a^kD^k}{k!} x^m = \sum_{k=0}^{\infty}\frac{\Gamma(m+1)}{\Gamma(k-m+1)k!} a^k x^{m-k} = \sum_{k=0}^{\infty}\frac{m!}{(k-m)!k!} a^k x^{m-k}$$ $$= \sum_{k=0}^{\infty}{m \choose k} a^k x^{m-k} = \sum_{k=0}^{m}{m \choose k} a^k x^{m-k} = (x+a)^m\,, $$ by binomial theorem. That means the differential operator works as a shift operator that shift the argument by $a$. Now, we can use this result to see the effect on a function $f(x)$. If the function $f(x)$ has a power series at $x=0$, then $$ {\rm e}^{aD} \sum_{k=0}^{\infty} f_k x^k = \sum_{k=0}^{\infty} f_k {\rm e}^{aD} x^k = \sum_{k=0}^{\infty} f_k (x+a)^k = f(x+a)\,.$$ So, the general effect of the differential operator ${\rm e}^{aD}$ on the function $f(x)$ is nothing but shifts the argument by $a$. In your case $a=1$.