Exponential of a function times derivative Exponential of a derivative $e^{a\partial}$ is simply a shift operator, i.e.
\begin{equation}
e^{a\partial}f(x)=f(a+x)
\end{equation}
This can be easily verified from a Taylor series
\begin{equation}
e^{a\partial}=\sum_{n=0}^\infty \frac{(a\partial)^n}{n!}
\end{equation}
and
\begin{equation}
f(x)=\sum_{n=0}^\infty c_n\frac{x^n}{n!}
\end{equation}
and applying one on another by using $\partial^mx^n=\frac{n!}{(n-m)!}x^{n-m}$ for $n\ge m$ and zero otherwise.
What if, instead of constant $a$, there is a function $g(x)$? In other words, I'm looking for $e^{g(x)\partial}f(x)$. Now, the derivative operator also acts on $g(x)$, which makes things very complicated and and seemingly intractable. For example, $[g(x)\partial]^2=g(x)[g'(x)+g(x)\partial]\partial$ and it gets worse for higher orders.
Also, where can I find the list of identities such as $e^{a\partial}f(x)=f(a+x)$? Searching for function (or exponential or logarithm) of a derivative gets clogged with results about derivative of a function (or exponential or logarithm).
 A: The question is answered affirmative (and in a much simpler way) elsewhere:

How to derive these Lie Series formulas

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:
$$
u(t) = e^{t\,g(x)\frac{d}{dx}} x \quad \Longleftrightarrow \quad \dot{u}(t) = g(u(t)) \quad \mbox{with} \quad x = u(0)
$$
A: 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:
Sophus Lie, Vorlesungen über Differentialgleichungen
mit bekannten Infinitesimalen Transformationen, bearbeitet und
herausgegeben von Dr. Georg Wilhelm Scheffers,Leipzig (1891). Availability:
Amazon ,
bol.com .

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.
