Exponential of a function times derivative
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:
$$
u(t) = e^{t\,g(x)\frac{d}{dx}} x \quad \Longleftrightarrow \quad \dot{u}(t) = g(u(t)) \quad \mbox{with} \quad x = u(0)
$$