Beltrami's Models
Beltrami's Models of Non-Euclidean Geometry by Nicola Arcozzi might be of interest. It does not start with the pseudosphere in the sense of the tractricoid, which is a finite surface of constant negative curvature embedded into three-dimensional Euclidean space. Instead, it describes planar models, one of which Arcozzi calls the projective model but which is known to me as the Beltrami-Klein model.
Quoting Arcozzi:
Beltrami calls psedudospheres the surfaces bijectively parametrized by the coordinates $(u,v)$ and endowed with the metric (1) (below, we will use idifferently [sic] psedosphere, hyperbolic plane, non-Euclidean plane)
So even though Beltrami started describing the tractricoid (and other surfaces of revolution with constant negative curvature, iirc), here he apparently is using the term in a different meaning, and keeping these two meanings apart is important.
So what Beltrami did was come up with a model: a way to translate terms of the axioms into geometric representations. Namely a hyperbolic point shall be modeled by a point inside a given disc (or other conic, at least in Klein's version), and a hyperbolic line shall be modeled by a segment of that disc. He also redefines metric, in particular he defines lengths (this is the equation (1) the above quotation refers to). He then shows that this model has all the properties of hyperbolic geometry.
So if his Euclidean geometry is consistent, then his model works, therefore it has the properties it demonstrate it should have, therefore hyperbolic geometry is consistent. Or formulated the other way round, if there was a problem with hyperbolic geometry, then there would be some problematic configuration in this model, and since he deduced that proper Euclidean geometry can not cause any such problems, this would imply that there can be no proper i.e. consistent Euclidean geometry either.
Greenberg
Greenberg's Euclidean and Non-Euclidean Geometries states that
Beltrami proved the relative consistency of hyperbolic geometry in 1868 using differential geometry (see The Pseudosphere, Chapter 10).
At first I read this as supporting your claim that Beltrami did use the tractricoid directly to prove that consistency. But reading that chapter 10, I'm not so sure any more. It starts by mentioning that the hyperbolic plane cannot be isometrically embedded, but a portion of it can.
So I guess that Beltrami might have recognized that he can carry the metric of the tractricoid over to a portion of the plane, and then extend it to the whole disk in the consistent way expressed by that equation (1) in Arcozzi's text. So the tractricoid would serve as a tool to demonstrate that the metric he chose is sane and relates to constant negative curvature, but the hyperbolic plane used for the consistency proof goes beyond the tractricoid.
Arcozzi again
Reading more of Arcozzi suggests a different interpretation, though:
On the other hand, he seems to worry that the surface with the metric (1) might not be considered wholly “real”, since it is not clear in which relation it stands with respect to Euclidean three space (the strictest measure of “reality”). Then, he will show that, after cutting pieces of it, the pseudosphere can be isometrically folded onto a “real” constant curvature surface in Euclidean space.
However, reading even further, one finds section 3.3 where Arcozzi speculates on how Beltrami might have thought of his geometry. The image presented there (at least the way I understand it) is more that of a 3D curved surface, quite like the tractricoid, rather than of a flat surface equipped with some strange artificial metric computation. However, due to differences in how things were done at the time, self-intersection apparently was little concern, and similar for the fact that only a limited portion was representable this way. Particularly after Beltrami had demonstrated the possibility of isometric motions.
Beltrami himself
Skimming Beltrami's Saggio di interpretazione della geometria Non-Euclidea myself, I recognize that equation for the distance element of the Beltrami-Klein model. It is indeed numbered (1) in his work as well. At first glance I see no reference to the pseudosphere at all, only references to constant negative curvature. I don't speak Italian, but here is what it has to say about pseudospheres (this is from a different version which lacks illustrations but was typeset in $\TeX$):
Per evitare circonlocuzioni ci permettiamo di denominare pseudosferiche le superficie di curvatura costante negativa, e di conservare il nome raggio alla costante $R$ da cui dipende il valore della loro curvatura.
By Stillwell's translation:
To avoid circumlocution, we call the surface of constant negative curvature pseudospherical, and we retain the term radius for the constant $R$ on which its curvature depends.
This in my opinion supports Arcozzi's view on how this term was used.
Best Answer
There is a general fact in Riemannian geometry which follows from "standard" existence and uniqueness theorems in the study of ODEs. Mainly, given any point $p$ in a Riemannian manifold $M$ and given any $v\in T_pM$, there is a unique geodesic defined for some open neighborhood of time around 0 starting at $p$ in the direction of $v$.
Assuming this, it's not too hard to show that you've found all the geodesics. You just need to verify that for any point $p$ in the hyperbolic plane and tangent vector, either the vector points up/down (so a vertical line geodesic has $v$ as tangent vector), or some semicircle through $p$ hitting the $x$-axis perpendicularly does.
Finally, to show there is a unique geodesic between any two points, one argues as follows:
Given $p$ and $q$, if $q$ and $p$ have the same $x$ coordinate, a vertical line passes through them. Further, no semicircle hitting the $x$-axis perpendicularly can go through both $p$ and $q$ since the semicircle is the graph of a function, so passes the vertical line test.
Next, given $p$ and $q$ with different $x$ coordinates, it's clear that a vertical line can't connect them, so we need only find all semicircles passing through both perpendicular to the $x$-axis. Algebraically, proceed by noting the equation of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and $r$. Plugging in $(x,y) = (p_1,p_2)$ and $(x,y) = (q_1,q_2)$ respectively and setting the two equal to each other (since they both equal $r^2$), one gets $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$
By simplifying this, one sees the $a^2$ cancels out and so one gets an equation linear in $a$, so $a$ is uniquely determined. (The assumption that $p_1\neq q_1$ means that the equation has the linear piece). Knowing $a$ easily implies that $r$ is uniquely determined (keeping in mind that $r \geq 0$). This shows that there is a unique semicircle perpendicular to the $x$-axis going through $p$ and $q$.