What does $\exp\left( ax\frac{d}{dx} \right)$ do on $\psi(x)$?

I had a professor who called this the "Dilation Operator", though I haven't it referred by that name anywhere else. You can compute it using a neat coordinate transformation. To start, you should know that:

$$\exp\left({a\frac{\text{d}}{\text{d}y}}\right)g(y) = \sum_{n=0}^\infty \frac{a^n}{n!}\frac{\text{d}^n}{\text{d}y^n}g(y) = g(y+a),$$ which follows quite simply from the definition of the Taylor Series.

The trick to evaluate the operator you have is to perform a coordinate substitution from $x$ to some $y$, such that $$x \frac{\text{d}}{\text{d} x} \longrightarrow \frac{\text{d}}{\text{d}y}$$

You should be able to see that if you choose $y = \ln x$, this would be true, so $$a \left(x\frac{\text{d}}{\text{d}x}\right) f(x) = a \left(\frac{\text{d}}{\text{d}y} \right) f(e^y),$$

and from this it's quite easy to show that $$\exp\left({a x\frac{\text{d}}{\text{d}x}}\right) f(x) = \exp\left({a \frac{\text{d}}{\text{d}y}}\right) f(e^y) = f(e^{y+a}) = f(e^a x),$$

and voila, there you have it!

Edit: Here's another (simpler?) way to show it, using the properties of the Euler Operator $\left(x \frac{\text{d}}{\text{d}x}\right)$, whose eigenfunctions are the monomials $x^n$, and eigenvalues $n$.

$$\left(x \frac{\text{d}}{\text{d}x}\right) x^n = n\,\, x^n \quad \quad \implies \quad \quad \left(x \frac{\text{d}}{\text{d}x}\right)^m x^n = n^m x^n$$

If you can write $\psi(x) = \sum_n c_n x^n$, and expand the exponential operator using its power series definition, then:

\begin{aligned} \exp\left(a \frac{\text{d}}{\text{d}x}\right) \psi(x) &= \sum_m \frac{a^m}{m!}\left(x \frac{\text{d}}{\text{d}x}\right)^m \sum_n c_n x^n\\ &= \sum_n c_n \underbrace{\sum_m \frac{a^m}{m!}n^m}_{\exp{an}} \,\, x^n\\ &= \sum_n c_n \left(e^a x\right)^n\\ &= \psi(e^a x) \end{aligned}

A short proof is to note that $$ \frac{\partial}{\partial \alpha} \left(\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)\right) = x \frac d{dx} \left(\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)\right) $$ and $$ \frac{\partial}{\partial \alpha} \psi(e^\alpha x) = x \frac d{dx} \psi(e^\alpha x) $$ so $\psi(e^\alpha x)$ satisfies the same 1st order ODE in $\alpha$ as does $\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)$ with the same initial condition.