= Solution
Here $dy/ds=1$, $dx/ds=u^2$, $du/ds=0$, so from initial parameter $\xi$ we get $y=s$, $x=\xi+\xi^2y$ and $u=\xi$. Solving for the branch agreeing with $u=x$ at $y=0$, and rationalizing the apparent division by $y$, gives
$$
\boxed{u(x,y)=\frac{2x}{1+\sqrt{1+4xy}},\qquad1+4xy>0.}
$$
This formula is smooth on that <open set>, which contains every point of the initial line. It satisfies $x=u+yu^2$ and $1+2yu=\sqrt{1+4xy}>0$. <Differentiation> gives $u_x=1/(1+2yu)$, $u_y=-u^2/(1+2yu)$, verifying the equation and the data.
The transport field is $(u^2,1)$, with normal component one on $y=0$, so the initial line is everywhere non-characteristic. This explains why neighborhoods cover the entire initial line, although a single fixed-width strip need not work for unbounded $x$.
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