The quickest way to get you started is to refer you to my article, reference [5] (a pdf) on
http://en.wikipedia.org/wiki/Squaring_the_circle
and then to the fourth edition (2008) of Marvin Jay Greenberg's book, which is reference [6].
I'm guessing what you want is Bolyai's construction, given a line and a point off the line, of the two rays through the point that are asymptotic to the line, one in each direction. When I wrote the article, I relied on an earlier edition of Marvin's book, along with The Foundations of Geometry and the Non-Euclidean Plane by George E. Martin, which has a nice little section at the very end. There is also, now, Geometry: Euclid and Beyond by Robin Hartshorne.
The most complete reference I know on constructions is in Russian, by Smogorshevski, other very helpful books by Kagan and by Nestorovich. Of course, at this point I have my own versions of it all.
Your answer is correct, except that in the example you gave, one has to adjoin far more numbers than you described in order to get to that ordered Pythagorean not-Euclidean field K.
That example is described on p.594 of the fourth edition of my book Euclidean and Non-Euclidean Geometries: Development and History (W.H. Freeman, 2008). There I called such a plane where parallel lines have a unique common perpendicular an HE-plane, abbreviating "halb elliptisch" from Pejas' classification article. Since you are only interested in the Archimedean case in order to get a natural metric once a unit segment is chosen, your HE-plane satisfies the acute angle hypothesis (by the Saccheri-Legendre theorem). You then describe an HE-plane accurately as the interior of a virtual circle in the affine plane over an ordered Archimedean Pythagorean not-Euclidean field K. An Archimedean ordered field is a subfield of R (up to isomorphism), so when you metrically complete it, you get R.
To make your argument completely rigorous, you would have to prove that the metric completion of an Archimedean H-plane is again an H-plane. Then, since it is complete, it must be either the real Euclidean or the real hyperbolic plane (in the HE-plane case it is the real hyperbolic). As I pointed out on p.594 of my book and as you indicated, if the line-circle axiom holds, then an Archimedean H-plane must be either Euclidean or hyperbolic, but its coordinate field could be any Euclidean subfield of R, such as the field of constructible numbers or the field of real algebraic numbers.
See also my March 2010 MONTHLY survey paper mentioned by Will Jagy entitled "Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries." Section 2 is all about Will Jagy's results about regular-polygoning circles in the hyperbolic plane (you can't always "square" them), and Section 3 is about the undecidability and consistency of elementary geometry.
If you email me at mjg0@pacbell.net I will send you my latest updating of that article.
(The terminology for all this is confusing. Yes, Janos Bolyai did introduce the term "absolute geometry" for the common part of real Euclidean and real hyperbolic geometries. I have argued - and it has generally been accepted by other writers - that "neutral geometry" is a better name, because one remains neutral about which parallel postulate to assume. I also argued that "absolute geometry" should be the name for Bachmann's geometry based on reflections, since it includes not only neutral geometry but also elliptic and other geometries - see p.588 of my book. Furthermore, even for neutral geometry, why restrict to real geometries? A model of Hilbert's incidence, betweenness and congruence axioms we now call a Hilbert plane or an H-plane for short. Pejas classified all H-planes. His classification is described on pp. 588ff of my book.)
Best Answer
I can do it if I am allowed to cheat - in axiom one for the set square (which is the axiom for the straight-edge) I'll allow one of the points to be ideal: that is, on the Gromov boundary of $H^2$. Also, I assume that the set square produces infinite geodesic rays. In particular, this means, in the second axiom, that the given point can be on the given line.
Suppose the segment $[x,y]$ is given. Draw the geodesic ray $P$, based at $x$ and perpendicular to $[x,y]$. Do the same at $y$ to get the ray $Q$, on the same side of $[x,y]$. Connect the ideal endpoint of $P$ to $y$ and the ideal endpoint of $Q$ to $x$. These new rays cross at $z$. Drop a perpendicular from $z$ to $[x,y]$ and we are done.