Utility of the operator $T^s=e^{s\frac{d}{dx}B(x)}$

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I came across this question: Is there a well-defined exponential of an integral operator?.

It just got me wondering about some elementary basics of operator theory. I think I understand basically what an operator is. I've worked with some important operators, such as integral transforms, the derivative and the integral.

My question is about the utility of the operator: $$ T^s=e^{s\frac{d}{dx}B(x)} $$

Is my understanding correct that this is (a) an operator and (b) the operator's input is $B(x)$ and the operator takes that input and spits out another function?

How is $T^s$ different than the shift operator? In what operator theory context does $T^s$ show up? Or does it not show up?

I computed an example. For $B(x):=x$ then $T^s=e^s.$ So monomials transform to exponentials under this operator.

Thank you for bettering my understanding here.

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$T^{s}\circ B=e^{s{d B(x)\over dx}}$ is indeed a (non linear) operator. The shift operator is different: $S^{s}\circ B=e^{d\over dx} B(x)$, that is, $S^{s}\circ B=(1+s{d\over dx}+{s^2\over 2}{d\over dx}^2\cdots) B(x)$ which is the Taylor expansion of $B(x+s)$. They don't look obviously connected.