= Solution
With parameter $t$ along a <characteristic curve>, the equations are
$$
\frac{dx}{dt}=y,\qquad\frac{dy}{dt}=1,\qquad\frac{du}{dt}=u,\qquad(x,y,u)|_{t=0}=(s,0,s).
$$
Their solutions are $y=t$, $x=s+t^2/2$ and $u=se^t$. Thus the projected curve satisfies $y^2=2(x-s)$, as required. Eliminating the parameters gives
$$
\boxed{u(x,y)=\left(x-\frac{y^2}2\right)e^y.}
$$
Indeed $u_x=e^y$ and $u_y=e^y(x-y^2/2-y)$, so $yu_x+u_y=u$, and the initial values hold.
The parameter map $(s,t)\mapsto(s+t^2/2,t)$ has Jacobian one and is globally invertible. Every point whose <characteristic curve> meets the initial segment therefore lies in the strip
$$
\boxed{1\leq x-\frac{y^2}2\leq2,\qquad -\infty<y<\infty.}
$$
The uniquely determined open region is its interior, with the boundary values determined on the two endpoint <characteristic curves>. Both signs of $y$ are included. Outside the strip, a general solution is $u=e^yF(x-y^2/2)$ with $F(s)=s$ only required for $1\leq s\leq2$. Smooth perturbations of $F$ supported outside that interval produce different solutions with the same prescribed data, proving the stated region is exact.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2006-characteristic-strip.png]
{title=The characteristic strip determined by the entire initial segment}
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