Exponential of a function times derivative

Instead of translation of a function $f(x) \to f(x+a)$ , let us consider scaling. This means that we are going to make intervals of the independent variable smaller, or larger, with a factor $\lambda > 0$. The transformed function is then defined by: $$ f_\lambda(x) = f(\lambda\,x) $$ Like with translations, it would be nice to develop the function $f_\lambda(x)$ into a Taylor series expansion around the original $f(x)$. But this is not as simple as in the former case. Unless some clever trick is devised, which reads as follows. Define a couple of new variables, $a$ and $y$, and a new function $\phi$ : $$ \lambda = e^a \qquad ; \qquad x = e^y \qquad ; \qquad \phi(y) = f(e^y) $$ Then, indeed, we can develop something into a Taylor series: $$ f_\lambda(x) = f(e^a\, e^y) = f(e^{a+y}) = \phi(y+a) = e^{a \frac{d}{dy}} \phi(y) $$ A variable such as $y$, which renders the transformation to be like a translation, is called a canonical variable. In the case of a scaling transformation, the canonical variable is obtained by taking the logarithm of the independent variable: $y = \ln(x)$. Working back to the original variables and the original function: $$ \phi(y) = f(e^y) = f(x) \qquad ; \qquad a = ln(\lambda) $$ $$ \frac{d}{dy} = \frac{dx}{dy}\frac{d}{dx} = e^y\frac{d}{dx} = x\frac{d}{dx} $$ Where the operator $x\,d/dx$ is called the infinitesimal operator of a scaling transformation. Such an infinitesimal operator always equals differentiation to the canonical variable, which converts the transformation into a translation. We have already met, of course, the infinitesimal operator for the translations themselves, which is simply given by $(d/dx)$. This leads rather quickly to the following: $$ f_\lambda(x) = f(\lambda\, x) = e^{\ln(\lambda)\, x\frac{d}{dx}} f(x) $$ Which is somewhat bogus, because of some artificial restrictions imposed by our heuristics: $x=e^y$ had to be positive, for example. So let's specify this for the scaling transformation of $x$ itself, which is represented by the series $\;e^{\ln(\lambda)\, x\, d/dx} x$ : $$ e^{ln(\lambda) \,x \frac{d}{dx}} x = x + \ln(\lambda)\, x\frac{dx}{dx} + \frac{1}{2} \ln^2(\lambda)\, x\frac{d(x\,dx/dx)}{dx} + \cdots $$ $$ = \left[1 + \ln(\lambda) + \frac{1}{2} \ln^2(\lambda) + \cdots \right] x = e^{\ln(\lambda)} x = \lambda\, x $$ Similarly (Update): $$ e^{ln(\lambda) \,x \frac{d}{dx}} x^n = x^n + \ln(\lambda)\, x\frac{dx^n}{dx} + \frac{1}{2} \ln^2(\lambda)\, x\frac{d(x\,dx^n/dx)}{dx} + \cdots \\ = x^n + \ln(\lambda)\,n\,x^n + \frac{1}{2} \ln^2(\lambda)\,n^2\,x^n + \frac{1}{6} \ln^3(\lambda)\,n^3\,x^n + \cdots \\ = \left[1 + \ln(\lambda^n) + \frac{1}{2} \ln^2(\lambda^n) + \frac{1}{6} \ln^3(\lambda^n) + \cdots \right] x^n = e^{\ln(\lambda^n)} x^n = \lambda^n\, x^n $$ Suppose that $f(x)$ can be written as a Taylor series expansion, then for all $x \in \mathbb{R}$ : $$ e^{ln(\lambda) \,x \frac{d}{dx}} \left[ a_0 + a_1 x + a_2 \frac{1}{2} x^2 + \cdots \right] = a_0 + a_1 (\lambda\,x) + a_2 \frac{1}{2} (\lambda\,x)^2 + \cdots \\ \Longrightarrow \qquad e^{ln(\lambda) \,x \frac{d}{dx}} f(x) = f(\lambda\,x) $$ (End of update) Since $\lambda$ must be positive, there exists no continuous transition towards problems where values are, at the same time, inverted or mirrored, like in: $$ f_\lambda(x) = f(-\lambda\,x) $$ For this to happen, the scaling transformation would to have to pass through a point where things are contracted to zero: $$ f_\lambda(x) = f(0\,x) $$ This already reveals a glimpse of the topological issues which may be associated with Lie Groups : remember that keyword. To be honest, I haven't seen any other generalization of your problem in 1-D, except the above scaling example.

Update. Well, not really. After some digging in my old notes, I've found a little bit more.
Consider the operation $\;e^\alpha\,x\;$ with $\;\alpha = g(x)\frac{d}{dx}$ . Then by definition: $$ e^{\alpha \,x} = 1 + \alpha\, x + \frac{1}{2} \alpha \left( \alpha\, x \right) + \frac{1}{3} \alpha \left( \frac{1}{2} \alpha \left( \alpha\, x \right)\right) + \cdots \\ $$ This can be written recursively as: $$ e^\alpha \, x = x + \alpha_1 x + \alpha_2 x + \alpha_3 x + \cdots \qquad ; \qquad\alpha_1 = \alpha \qquad ; \qquad \alpha_n = \frac{1}{n} \alpha \, \alpha_{n-1} $$ We have seen cases where $\;g(x) = a\;$ and $\;g(x) = \ln(\lambda)\,x$ . Now let's try another example, with $\;g(x) = x^2$ : $$ \alpha_1 x = x^2 \frac{d}{dx} x = x^2 \\ \alpha_2 x = \frac{1}{2} x^2 \frac{d}{dx} x^2 = x^3 \\ \alpha_3 x = \frac{1}{3} x^2 \frac{d}{dx} x^3 = x^4 \\ \cdots \\ \alpha_n = \frac{1}{n} x^2 \frac{d}{dx} x^n = x^{n+1} \\ $$ Consequently, say for real $0 < x < 1$ : $$ e^{x^2\,d/dx} x = x + x^2 + x^3 + \cdots + x^n + \cdots = \frac{x}{1-x} $$ Which can perhaps be generalized to functions $f(x)$ that have a Taylor expansion.
One might think now that the above results may be combined as follows: $$ e^{(ax^2+bx+c)d/dx}x = e^{c\,d/dx}e^{bx\,d/dx}e^{ax^2\,d/dx}x = e^{ax^2\,d/dx}e^{bx\,d/dx}e^{c\,d/dx}x $$ But it can readily be verified that such is not the case. The reason is that the operators $\;x^2\,d/dx$ , $x\,d/dx$ , $d/dx\;$ do not commute. Define the commutator $\left[\,,\right]$ of two operators $\alpha$ and $\beta$ as: $$\left[\alpha\,,\beta\right] = \alpha\beta - \beta\alpha$$ Then prove that: $$ \left[ x^2\frac{d}{dx}, x\frac{d}{dx} \right] \ne 0 \qquad ; \qquad \left[ x^2\frac{d}{dx}, \frac{d}{dx} \right] \ne 0 \qquad ; \qquad \left[ x \frac{d}{dx} , \frac{d}{dx} \right] \ne 0 $$

Late revision. I've ordered the following book and reading it now:

Formulated in somewhat outdated notation I find the following IMHO astonishing Theorem on page 50 and next. Operator notation is mine: $$ \boxed{ \; e^{t \phi(x) \frac{d}{dx}} f(x) = f\left(e^{t \phi(x) \frac{d}{dx}} x\right) \; } $$ Here $\phi(x)$ and $f(x)$ are "neat" but for the rest quite arbitrary functions. Therefore the differential operator and the function are always commutative, which is quite a non-trivial fact. When applied to the last of the above examples (slightly modified) we have via page 75 of the book: $$ e^{t x^2 \frac{d}{dx}} f(x) = f\left(\frac{x}{1-x t}\right) $$ So it is indeed sufficient to apply the operator $\exp(t \phi(x) d/dx)$ to the independent variable $x$ only. If that results in a closed form, then you can apply the Theorem and have a closed form for any other function $f(x)$ as well.

Sad remark. The book by Georg Scheffers is abundant with "non rigorous" notions, especially infinitesimals. The latter are quite essential for understanding the book. For me, as a physicist by education, this represents no problem at all. But I know from bad experience that those good old infinitesimals represent sort of a taboo for modern mathematics. Therefore, in retrospect, it can be understood very well why this approach by Georg Scheffers hasn't found wide audience among professional mathematicians. Even worse. I find that professional mathematicians rather have distorted the original theory as meant by Sophus Lie a great deal. Such that essential parts of it, like the above Theorem, tend to be erased from common mathematical knowledge. Which I hope not.

The question is answered affirmative (and in a much simpler way) elsewhere: Summary. First solve the differential equation: $$ g(x) = \frac{1}{\phi'(x)} \quad \Longrightarrow \quad \phi(x) = \int \frac{dx}{g(x)} $$ Then we have (barring division by zero and other issues): $$ e^{g(x)\partial} f(x) = f(\phi^{-1}(\phi(x)+1)) $$ Update, triggered by another question (but where?)
By definition, for a function $\phi$ and its inverse: $$ y = \phi(x) \quad \Longleftrightarrow \quad x = \phi^{-1}(y) \quad \Longleftrightarrow \quad \phi^{-1}(\phi(x)) = x $$ From this, an elementary result in calculus follows: $$ \frac{dy}{dx} \frac{dx}{dy} = 1 = \frac{d\phi(x)}{dx} \frac{d\phi^{-1}(y)}{dy} \quad \Longrightarrow \\ \frac{d\phi^{-1}(y)}{dy} = \frac{1}{\phi'(x)} = \frac{1}{\phi'(\phi^{-1}(y))} \quad \Longrightarrow \\ \frac{d\phi^{-1}(x)}{dx} = \frac{1}{\phi'(\phi^{-1}(x))} $$ There is an application with the Lie series. We have: $$ u(t) = e^{t\,g(x)\frac{d}{dx}} x = \phi^{-1}(\phi(x)+t) \quad \mbox{with} \quad g(x) = \frac{1}{\phi'(x)} \\ u(0) = e^{0\,g(x)\frac{d}{dx}} x = x = \phi^{-1}(\phi(x)) $$ It follows that: $$ \frac{du}{dt} = \frac{d\phi^{-1}(\phi(x)+t)}{dt} = \frac{1}{\phi'(\phi^{-1}(\phi(x)+t))} = \frac{1}{\phi'(u(t))} $$ In short: $$ u(t) = e^{t\,g(x)\frac{d}{dx}} x \quad \Longleftrightarrow \quad \dot{u}(t) = g(u(t)) \quad \mbox{with} \quad x = u(0) $$