= Solution
Choose one <projective line> from each of two different pairs. They are <skew lines> by the preceding disjointness result. After a projective linear coordinate change call them
$$
L=\{x_2=x_3=0\},\qquad M=\{x_0=x_1=0\}.
$$
For $A\in L$, $B\in M$, their joining line meets the <cubic surface> at $A,B$ and a third point. This defines a <rational map> $L\times M\dashrightarrow S$ on the <Zariski-open set> where that third intersection is distinct and the joining line is not contained in $S$.
More explicitly, because $F$ vanishes on both <projective lines>,
$$
F(\alpha A+\beta B)=\alpha\beta\bigl(\alpha U(A,B)+\beta V(A,B)\bigr).
$$
The third point is $[VA-UB]$. Both <polynomials> $U,V$ are nonzero: if either vanished identically, one of the two <projective lines> would lie in the <singular locus>. Thus $UV\ne0$ supplies a nonempty <Zariski-open set>.
Conversely, a point $P=[x_0:x_1:x_2:x_3]$ outside $L\cup M$ lies on the unique joining line whose endpoints have coordinates $[x_0:x_1]$ on $L$ and $[x_2:x_3]$ on $M$. This is a rational inverse on the same generic locus. The <rational parametrization of a cubic surface from two skew lines> gives
$$
S\ \text{birational to}\ L\times M\cong\mathbb P^1\times\mathbb P^1.
$$
The latter contains the <affine plane> as a dense open subset, as does $\mathbb P^2$. Hence $S$ is <birational> to $\mathbb P^2$.
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