Let $x(u,v)$ be such a parametrisation. We defined $$E = x_u \cdot x_u, \quad F = x_u \cdot x_v, \quad G = x_v \cdot x_v$$ Since these are 1, 0, 1 respectively, $x_u$ and $x_v$ are always orthonormal and so the associated unit normal is $\xi := x_u \times x_v$. Then we define $$l = x_{uu} \cdot \xi, \quad m = x_{uv} \cdot \xi,\quad n = x_{vv} \cdot \xi$$ and these are also 1, 0, 1.
Differentiating $x_u \cdot x_u = 1$ wrt $u$ and $v$ yields us $x_{uu} \cdot x_u = 0$ and $x_{uv} \cdot x_u = 0$.
Differentiating $x_u \cdot x_v = 1$ wrt $u$ yields $$x_{uu} \cdot x_v = - x_u \cdot x_{uv} = 0.$$ So we find that $x_{uu}$ is orthonormal to the surface. Same thing with $x_{vv}.$
So $$x_{uu} = (x_{uu} \cdot \xi) \xi = l \xi = \xi$$ and similarly $x_{vv} = \xi$ and $x_{uv} = 0$. But this is where I get stuck. I am not able to find a contradiction and also cannot find an example.
Using an expression we learned for the shape operator $S$ in function of $E,F,G,l,m,n$, and using $\xi_u = -S x_u$ I also found that $\xi_u = - x_u$ and similarly $\xi_v = - x_v$. So it might work if we would take the unit sphere. (Since the position vector $x$ is also the unit normal, modulo the sign.) But I can’t find the example. :(
Any help is appreciated!
I believe I have solved this problem. If anyone wants to critique the answer, please go ahead!
Say the matrix form of the Shape operator in the basis $(x_u, x_v)$ is given by $$ S x_u = a x_u + c x_v, \quad S x_v = b x_u + d x_v.$$ I know from class that $S x_u = - \xi_u$ and $S x_v = - \xi_v$. So we find $$a = (S x_u) \cdot x_u = - \xi_u \cdot x_u = \xi \cdot x_{uu} = l = 1.$$ In similar fashion we can find that $b = c = 0$ and $d = 1$. Hence, the Shape operator is the identity.
However, since $x_{uu} = x_{vv} = \xi$and $x_{uv} = 0$ from earlier, we also have $$S x_v = - \xi_v = x_{uuv} = (x_{uv})_u = 0_u= 0,$$ (and similarly $S x_u = 0). This contradicts the fact that the shape operator is the identity.
Hence, such an $x$ which we had presumed to exist does not exist.