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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 themFor $A\in L$, $B\in M$, their joining line meets the cubic surface at $A,B$ and a third point. This defines a rational map $L\times M⇢S$ 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,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 $UV\ne 0$ 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 givesThe latter contains the affine plane as a dense open subset, as does ${\mathbb{P}}^2$. Hence $S$ is birational to ${\mathbb{P}}^2$.