Let us recall some basic ideas. Define
$ \hat{f}(\xi) = \int_{\Bbb{R}} f(x)e^{-2\pi i \xi x} \, dx \quad \text{and} \quad \check{f}(x) = \int_{\Bbb{R}} f(\xi)e^{2\pi i x \xi} \, d\xi. $
Then for a function $f \in \mathcal{S}(\Bbb{R})$ of Schwartz class, we have
$ \left(\frac{d^n}{dx^n} f \right)^{\wedge}(\xi) = (2\pi i \xi)^n \hat{f} (\xi). $
Thus if we put $D = \frac{1}{2\pi i} \frac{d}{dx}$, we have
$ D^n f(x) = \big(\xi^n \hat{f}(\xi)\big)^{\vee}(x).$
That is, under Fourier transform, differentiation operator is indeed a multiplicative operator. This observation is the key ingredient to define a general class of differential operators. Indeed, for $g \in L^{\infty}(\Bbb{R})$ the differential operator
$g(D) : L^2(\Bbb{R}) \to L^2(\Bbb{R})$
is define as
$ g(D)f(x) = \big(g(\xi) \hat{f}(\xi)\big)^{\vee}(x), \qquad f \in L^2(\Bbb{R}). $
This definition completely make sense since $g(\xi)\hat{f}(\xi) \in L^2(\Bbb{R})$ by Hölder's inequality. (If we change the condition of $g$, then the corresponding operator $g(D)$ no longer stands for an endomorphism on $L^2(\Bbb{R})$, but rather an operator between two Sobolev spaces.)
For example, we can formally write $e^{-s \frac{d}{dx}} = g(D)$ where $g(x) = e^{-2\pi i s x} \in L^{\infty}(\Bbb{R})$. Then by utilizing some identities, we have
$ e^{-s \frac{d}{dx}}f(x) = g(D)f(x) = \big(e^{-2\pi i s \xi} \hat{f}(\xi)\big)^{\vee}(x) = f(x-s).$
If you want to replace $D$ by some other differential operators, you may instead change $g$ by some other function.