We don't even need Christoffel symbols or curvature:
We may assume that $S$ is given in the form ${\bf f}: \quad (x,y)\mapsto \bigl(x,y, f(x,y)\bigl)\qquad(*)$ with $f(x,0)\equiv0$, $f(0,y)\equiv0$, i.e., the $x$-axis and the $y$-axis are lines on $S$.
By assumption through each point $(x,0,0)$ there is a line $g_x$ on $S$ which is orthogonal to the $x$-axis; therefore $g_x$ can be written in the form ${\bf g}_x:\quad s\mapsto\bigl(x, s, p(x) s\bigr)\ ,$ where $p(x)$ denotes the slope of $g_x$ with respect to the $(x,y)$-plane.
Similarly, through each point $(0,y,0)$ there is a line $h_y$ on $S$ which is orthogonal to the $y$-axis; therefore $h_y$ can be written in the form ${\bf h}_y:\quad t\mapsto\bigl(t, y, q(y) t\bigr)\ ,$ where again $q(y)$ denotes the slope of $h_y$ with respect to the $(x,y)$-plane.
The two lines $g_x$ and $h_y$ intersect at the point ${\bf r}_{xy}\in S$ characterized by ${\bf g}_x(y)={\bf h}_y(x)$. Looking at the third coordinate of ${\bf r}_{xy}$ we see that $p(x)y=q(y) x\ .$ As this is true for all $x$ and $y$ near $0$ there has to be a constant $c\in{\mathbb R}$ with $p(x)=cx$, $q(y)=cy$. This implies that the $f$ in $(*)$ is given by $f(x,y)=c\>xy$, or that $S$ is a parabolic hyperboloid, when $c\ne0$.
Since $g_x$ and $h_y$ should intersect orthogonally at ${\bf r}_{xy}$ we must have $0={\bf g}_x'(y)\cdot {\bf h}_y'(x)=(0,1, cx)\cdot(1,0, cy)=c^2 x y\ ,$ which implies $c=0$, so that in fact $f(x,y)\equiv0$, and $S$ is a plane.