The question was originally asked here Doubly Ruled Surfaces and I am following the hint provided by the OP. That is, first show that $K\equiv0$ and then deduce that the surface is a plane.
Let the surface be $x(u,v)$. If we suppose w.l.o.g. that locally the orthogonal lines are $u$- and $v$-curves, then we have $F\equiv 0$ and we know that $K=-\frac{1}{2\sqrt{EG}}\Big(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v+\Big(\frac{G_u}{\sqrt{EG}}\Big)_u\Big)$. I am confused about how to use the fact that the parameter curves are straight lines. Do we have $x_{uu}\cdot x_v=0$ which would imply $E_v=0$ and similarly, $x_{vv}\cdot x_u=0$ which would imply $G_u=0$?
If we know that $K\equiv 0$, how do we deduce that the surface is a plane?
This question looks familiar. Yes, your approach to show $K=0$ is correct. Now remember that a plane is characterized by having $0$ shape operator (or $0$ second fundamental form), so see what you can figure out about the second fundamental form.