Why is wave drag non-zero for a flow without shockwaves?
The drag can be computed from Navier-Stokes momentum equations as:

$$D=-\oint_{S_\infty}p\hat{i}\cdot d\vec{S} - \oint_{S_\infty}\rho (\hat{i}\cdot\vec{u})(\vec{u}\cdot d\vec{S})$$

Usually, all boundaries other than the rear boundary (Trefftz plane) would have the same flow properties as the free-stream, and they fall out of the integral. We are left with:

$$D=\int_{S_T}{\rho u(V_\infty-u)dS}$$

As you correctly pointed out, the momentum defect of a potential flow should be exactly zero, which means that drag should also be zero.

However, for oblique shock in small-disturbance potential flow, the oblique shock (or rather, each constant potential wave characteristic) extends to infinity, carrying energy with it. If you want to calculate the far-field drag, you would need to do the integral around the entire far-field. This is different from energy dissipation in viscous boundary layer or subsonic normal shock. 



Picture ref. https://nikander.github.io/compflow/Anderson/Chapter4/print/