I think that the proof can be given directly, which gives it more clarity, using Propositions I.13, I.17, I.28 and I.31 (which do not depend on the Fifth Postulate), following the demonstration in the book Introduction to non-Euclidean geometry by Harold E. Wolfe (Chapter 2).
First, note that the line parallel to a line given by a point not on the line constructed in Proposition I.31 ("To draw a straight line through a given point parallel to a given straight line") produces internal angles of the same side. whose sum is equal to two right angles, this is deduced by virtue of Proposition I.13 ("If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles").
Now, to prove that Playfair Axiom implies the Fifth Postulate it is as follows:
Given lines $AB$ and $CD$ cut by the transversal $ST$ in such a way that the sum of angles $BST$ and $DTS$ is less than two right angles. Construct through $S$ the line $QSR$, making the sum of angles $RST$ and $DTS$ equal to two right angles. This line is parallel to $CD$ by Proposition I.28 ("If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another"). Since lines $QSR$ and $ASB$ are different lines and, by Playfair’s Axiom, only one line can be drawn through $S$ parallel to $CD$, we conclude that $AB$ meets $CD$. These lines meet in the direction of $B$ and $D$, for, if they met in the opposite direction, a triangle would be formed with the sum of two angles greater than two right angles, contrary to Proposition I.17 ("In any triangle the sum of any two angles is less than two right angles").