I feel like I am close to piecing together the relation but am not quite fully sure.
This is the shift operator I’m talking about from Wikipedia page on shift operator
The shift operator $T^{t}$ (where $t \in \mathbf{R}$ ) takes a function $f$ on $\mathbf{R}$ to its translation $f_{t}$, $T^{t} f(x)=f_{t}(x)=f(x+t)$. A practical operational calculus representation of the linear operator $T^{t}$ in terms of the plain derivative $\frac{d}{d x}$ was introduced by Lagrange, $$T^{t}=e^{t\frac{d}{d x}}, $$ which may be interpreted operationally through its formal Taylor expansion in $t ;$ and whose action on the monomial $x^{n}$ is evident by the binomial theorem, and hence on all series in $x$, and so all functions $f(x)$ as above. $[3]$ This, then, is a formal encoding of the Taylor expansion in Heaviside's calculus.
I also know the properties of Fourier transform time and frequency shifting from the page on the Fourier transform
Translation / time shifting
For any real number $x_{0}$, if $h(x)=f\left(x-x_{0}\right)$, then $\hat{h}(\xi)=e^{-2 \pi i x_{0} \xi} \hat{f}(\xi)$.
Then heuristically because of my slight knowledge of operational calculus, I know that the derivative operator with respect to time can be equated to frequency.
Conversely because of the duality of time and frequency, I’m guessing the derivative operator with respect to frequency is the same as the time variable.
I feel like there should be some way to interpret all of these things in a unified manner but I haven’t quite figured it out. Mainly the derivative shift operator applied to a function looks like the Fourier transform shifting but there is a slight difference between the two when you do the calculations.
From the Fourier side you usually speak of a Fourier multiplier $m(\partial/i)$ as the operator $$ \widehat{m(\partial/i)f}(\xi):=m(\xi)\hat f(\xi).$$ This is to match with e.g. $\widehat{\partial f}=+i\xi\hat f$. Equating $e^{s\partial}=m(\partial/i)$ we get $m(\xi)=e^{is\xi}$ and taking the inverse Fourier transform, we recover the translation operator as in the operational calculus: $$ (e^{s\partial}f)(x)=\int_{\mathbb R}e^{is\xi} \hat f(\xi)e^{ix\xi}d\xi = \int_{\mathbb R} \hat f(\xi)e^{i(x+s)\xi}d\xi =\overset{\vee}{\hat f}(x+s)=f(x+s). $$