Here's a sketch of an argument for approach 2, under the mild hypothesis that the lines in the family of doubly ruled lines varies continuously, which I think is intuitively clear, but requires some justification.

Take a point $p$ on a doubly ruled surface $\Sigma$, and take two lines $l_1, l_2$ going through $p$. For any nearby points $p_1 \in l_1, p_2 \in l_2$, there are lines $l_2'$ intersecting $l_1$ in $p_1$, and $l_1'$ intersecting $l_2$ in $p_2$. But since $l_1$ and $l_2$ intersect on the surface $\Sigma$, and since the family varies continuously by hypothesis, $l_1'$ and $l_2'$ must intersect when $p_1,p_2$ are close enough to $p$ by general position on $\Sigma$. Similarly, for any point $q$ near $p$, there are lines $l_1'$ and $l_2'$ going through $q$ meeting $l_2$ and $l_1$ respectively by continuity of the family and general position on $\Sigma$.

Now, take points $p_1, p_1'$ on $l_1$ near $p$, together with lines $l_2'$ and $l_2''$ meeting $l_1$ in $p_1$ and $p_1'$ respectively. Then $l_1'$ must meet both $l_2'$ and $l_2''$ for points $p_2$ near $p$ on $l_2$. Thus, one sees a neighborhood $p\in U\subset \Sigma$ such that any point in $U$ lies on a line meeting these three lines $l_2,l_2',l_2''$.

If any pair of these lines is coplanar (i.e. intersect or are parallel), then the portion of surface $U$ near $p$ must be planar. Otherwise, one has 3 skew lines.

Now, I claim that for 3 skew lines, there is a unique surface of lines meeting all three lines, which is either a hyperbolic paraboloid or hyperboloid. This is proved by Hilbert-Cohn Vossen, but I'll give a sketch of the proof. The uniqueness follows because for any point $p$ in $l_2$, the plane spanned by $p$ and $l_2'$ meets $l_2''$ uniquely in a point $q$, and thus every point on $l_2$ goes through a unique line $\overline{pq}$ meeting $l_2'$ and $l_2''$.

Three skew lines in $\mathbb{R}^3$ give three projective lines in $\mathbb{RP}^3$ which do not intersect. I claim that this configuration is unique up to projective transformation.

Three skew projective lines in $\mathbb{RP}^3$ correspond to three planes $P_1,P_2,P_3$ in $\mathbb{R}^4$ meeting pairwise only in the origin. I claim $GL_4(\mathbb{R})$ acts transitively on such configurations. Take basis vectors for $P_1, P_2$, then these form a basis for $\mathbb{R}^4$. Thus up to linear transformation, we may assume $P_1=\{(1,0,0,0),(0,1,0,0)\}, P_2=\{(0,0,1,0),(0,0,0,1)\}$. Now, the subspace $P_3$ has a basis of two vectors, such that the first two coordinates are linearly independent in $P_1$, and the second two coordinates are linearly independent in $P_2$ since we assume that the planes intersect only in the origin. We may take linear transformations in $GL_2(\mathbb{R})\times GL_2(\mathbb{R})$ stabilizing $P_1, P_2$ and sending these two vectors to $\{(1,0,1,0),(0,1,0,1)\}$, and thus we have normalized the three planes, so that the action is transitive. The corresponding action of $PGL_4(\mathbb{R})$ is thus also transitive on skew projective lines.

Take a piece of a hyperbolic paraboloid, and take 3 skew lines lying on it. The three skew lines uniquely determine the paraboloid, since it is the surface of lines meeting the three skew lines. Then there is a projective transformation taking $l_2,l_2',l_2''$ to these three skew lines, and therefore sending a portion of $\Sigma$ near $p$ to a hyperbolic paraboloid by uniqueness.

In fact, when we compactify a hyperbolic paraboloid or a hyperboloid in $\mathbb{RP}^3$, we get a 2-torus with two foliations by projective lines. So up to projective transformation, there is only one such surface.

Now the surface $\Sigma$ has patches which are hyperbolic paraboloids or hyperboloids. But one can see that each such surface lies in a unique doubly ruled torus in $\mathbb{RP}^3$, so $\Sigma$ must be identically such a surface intersected with $\mathbb{R}^3$.