This is an exercise in elementlary differential geometry (named as Catalan's theorem). Though there are many proofs of this problem, I meet some trouble to prove it. My idea is as follows:
(1) Assume the surface is $\mathbf{r}(u,v)=\mathbf{a}(u)+v\mathbf{b}(u)$, here $|\mathbf{b}|=1$ and $\langle\mathbf{a}',\mathbf{b}\rangle=0$, $u$ is the arc length of $\mathbf{a}(u)$. Then $F=0$, $N=0$, $G=1$.
(2) By the definition of the mean curvature $H=\frac12\frac{LG-2MF+NG}{EG-F^2}$, we get $L=0$, i.e. $P(v)=(\mathbf{a}''+v\mathbf{b}'',\mathbf{a}'+v\mathbf{b}',\mathbf{b})\equiv0$. Hence all the coefficients of the quadratic polynomial $P(v)$ equal $0$. Especially, $(\mathbf{b}'',\mathbf{b}',\mathbf{b})=0$, $(\mathbf{a}'',\mathbf{a}',\mathbf{b})=0$.
(3) From $(\mathbf{b}'',\mathbf{b}',\mathbf{b})=0$, we know $\mathbf{b}$ is a plane curve. Let $\mathbf{b}(u)=(\cos u,\sin u,0)$.
(4) Let $\{\mathbf{a};\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$ be the Frenet frame of the curve $\mathbf{a}$. Then $\mathbf{a}'=\mathbf{e}_1$, $\mathbf{a}''=\mathbf{e}_1'=\kappa\mathbf{e}_2$, $\kappa$ is the curvature of $\mathbf{a}$. By $(\mathbf{a}'',\mathbf{a}',\mathbf{b})=0$, we have $$-\kappa\langle\mathbf{e}_3,\mathbf{b}\rangle=0. $$
(5) If $\kappa(u)\neq0$, then there exists a small interval $(u_1,u_2)$, such that $\langle\mathbf{e}_3,\mathbf{b}\rangle=0$. From this equation, one easily obtain the conclusion.
My question: How to deal with the case $\kappa(u)=0$?
If $\kappa(u)=0\ \ \forall u$, then $\mathbf{e}_1'=0,\ \mathbf{a}'=\mathbf{e}_1=C$, $C$ is some constant. Therefore $\mathbf{a}$ is a straight line.