How to solve this integral with transformation to polar coordinates?

First off, $D$ is described in polar coordinates by $1\leq r\leq e$, and $0\leq \phi\leq 2\pi$. That's the easy part. You read the bounds for $r$ straight off the inequality for $x$ and $y$ (that inequality literally says $1\leq r^2\leq e^2$). And $D$ goes all the way around the origin, so $\phi$ goes from $0$ to $2\pi$. $D$ is rotationally symmetric, and bounded by circles centered at the origin, so this is as easy as it gets in polar coordinates.

The slightly more tricky part is that $dx\,dy$ does not just become $dr\,d\phi$. Just like in the one-dimensional case, there is an additional factor appearing here when doing substitution, related to the derivatives of one variable expressed as a function of the other.

In this case, $dx\,dy$ becomes $r\,dr\,d\phi$. So what you want is $$ \int_1^e\left(\int_0^{2\pi}\frac{\ln(r^2)}{r^2}r\,d\phi \right)dr\\ = 2\pi\int_1^e\frac{2\ln(r)}{r}\,dr $$ (where the inner $\phi$ integral disappears because the integrand does not depend on $\phi$).

First draw your "D" in $xy$ plane. It is clearly annulus shaped. The region is basically between two circles of radius $1$ and $e$. This is your limits for $r$. To cover entire annulus part, we need $0\leq \phi \leq 2\pi$.