= Solution
For fixed $t$, eliminate $s=x-t$ to obtain the projected <parabola>
$$
\boxed{y=x-2t+(x-t)^2.}
$$
Along that curve,
$$
x+y+1=(s+1)^2\geq0.
$$
Equality occurs at $s=-1$, where the point is $(t-1,-t)$. Also $dy/dx=y_s/x_s=1+2s=-1$ there, so the <tangent line> has the same slope as the line $x+y=-1$ and passes through a point of that line. Thus every projected characteristic is tangent to that line, with no crossing of it. Because $t\geq0$, the attained tangency points form only the half-line $x\geq-1$, $y=-x-1$.
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