The following theorem is definitely true in Euclidean geometry:
Theorem. Given any three parallel (that is, pairwise non-intersecting) lines $\ell_1, \ell_2, \ell_3$, there is a fourth line which intersects all three of them.
To prove it, pick a fourth line $\ell$ distinct from $\ell_1$ and intersecting $\ell_1$ at a point $P$. Then $\ell$ cannot be parallel to $\ell_2$ because there is only one line parallel to $\ell_2$ through the point $P$: the line $\ell_1$. So $\ell$ must intersect $\ell_2$. Similarly, $\ell$ must intersect $\ell_3$.
However, it is false in hyperbolic geometry. In the Poincaré half-plane model, let $\ell_1, \ell_2, \ell_3$ be the lines represented respectively by the semicircles with (ideal) endpoints $(0,0)$ and $(1,0)$; $(2,0)$ and $(3,0)$; $(4,0)$ and $(5,0)$. For a line to intersect $\ell_1$, it must have exactly ideal endpoint at $(x,0)$ for some $x \in (0,1)$; similarly it must have one with $x \in (2,3)$ for $\ell_2$ and $x \in (4,5)$ for $\ell_3$. All three of these cannot hold simultaneously.
Therefore, if we take the theorem above as an axiom, the parallel postulate must follow as a consequence.
Question. Is there a proof of this that does not rely on the "meta-reasoning" that all neutral geometries are either Euclidean or hyperbolic?
For concreteness, let's say we are working with Hilbert's axioms of plane geometry, without the parallel postulate (so, this list on Wikipedia, skipping I.4 through I.8, ignoring all mentions of "planes" elsewhere, and skipping IV.1). But I am not attached to this particular framework, as long as we're using an axiomatic system in which this question still makes sense. (We should pick something that rules out elliptic geometry, in which parallel lines don't exist at all.)