The answer is yes. The completion of a semi-elliptic plane is hyperbolic. The name probably comes from the fact that each pair of lines in it has either a point or a single perpendicular in common (in the elliptic plane they have both), but there is nothing elliptic about it metrically. Semi-elliptic planes are just hyperbolic planes thinned out so cleverly that no pair of asymptotic parallels remains. What of the paradox of removing all asymptotic parallels to a line and having nothing left? The catch is that some lines are not removed entirely. All their points are removed except one. It takes two points to form a line in geometry, so the plane does not "see" the lines that only have one point in it. What's unexpected is that most (uncountably many) lines are removed entirely and still the set of sole survivors is enough to support a geometry, in fact it is dense in the hyperbolic plane!

This is easiest to explain using the Klein model of a hyperbolic plane as the interior of a circle $D$ in $\mathbb{R}^2$. The lines are the chords of the circle, and the asymptotic parallels intersect on the boundary $\partial D$. Let $K$ be a proper subfield of $\mathbb{R}$ such that $K^2\cap\partial D=\emptyset$. Any two lines in $K^2$ intersect at a point with $K$ coordinates, so any line in $\mathbb{R}^2$ that intersects a $K^2$ line at a point in $\partial D$ has at most one point with $K$ coordinates. Everything so far works even with $K=\mathbb{Q}$, the rest in the construction of $K$ is to ensure that it is Pythagorean ($a,b\in K$ implies $\sqrt{a^2+b^2}\in K$), and that $K^2\cap\partial D=\emptyset$. Then $K^2\cap D$ is a semi-elliptic plane, and every semi-elliptic plane arises in this way. The metric is up to scale the restriction of the Klein-Beltramy metric on $D$ (logarithm of the cross-ratio), so the completion is the hyperbolic plane $D$ itself. Since according to Pejas's classification every Archimedean H-plane is Euclidean, hyperbolic or semi-elliptic (Theorem 3 in Greenberg's paper) the completion is always Euclidean or hyperbolic.

Since $K$ is only Pythagorean but not Euclidean (every positive element has a square root) there are lines in $K^2\cap D$ that pass through interior of circles without intersecting them in violation of the so-called line-circle axiom. This can also be seen geometrically. In any Archimedean H-plane that satisfies the hypothesis of the acute angle and the line-circle axiom a Bolyai's construction produces a line asymptotically parallel to any given one, which can not happen in $K^2\cap D$. The "metric constant" $k\in K$ is the Gaussian curvature of the completion, and since $\sqrt{-k}\notin K$ it can not always be normalized. In other words, not every hyperbolic plane can be thinned out into a semi-elliptic one, the standard one can't be among others. Pejas's example of $K$ is obtained by adjoining to $\mathbb{Q}(\sqrt{2})$ the square roots of all elements $x=a+b\sqrt{2}$ that are positive along with their conjugates $\overline{x}=a-b\sqrt{2}$, and taking a metric constant $k<0$ with $\overline{k}>0$, for instance $k=1-\sqrt{2}$. The equation of $D$ is then $x^2 +y^2<-k^{-1}$.

I'd like to thank Will Jagy for pointing me in the right direction in his previous answer

The answer is yes. The completion of a semi-elliptic plane is hyperbolic. The name probably comes from the fact that each pair of lines in it has either a point or a single perpendicular in common (in the elliptic plane they have both), but there is nothing elliptic about it metrically. Semi-elliptic planes are just hyperbolic planes thinned out so cleverly that no pair of asymptotic parallels remains. What of the paradox of removing all asymptotic parallels to a line and having nothing left? The catch is that some lines are not removed entirely. All their points are removed except one. It takes two points to form a line in geometry, so the plane does not "see" the lines that only have one point in it. What's unexpected is that most (uncountably many) lines are removed entirely and still the set of sole survivors is enough to support a geometry, in fact it is dense in the hyperbolic plane!

This is easiest to explain using the Klein model of a hyperbolic plane as the interior of a circle $D$ in $\mathbb{R}^2$. The lines are the chords of the circle, and the asymptotic parallels intersect on the boundary $\partial D$. Let $K$ be a proper subfield of $\mathbb{R}$ such that $K^2\cap\partial D=\emptyset$. Any two lines in $K^2$ intersect at a point with $K$ coordinates, so any line in $\mathbb{R}^2$ that intersects a $K^2$ line at a point in $\partial D$ has at most one point with $K$ coordinates. Everything so far works even with $K=\mathbb{Q}$, the rest in the construction of $K$ is to ensure that it is Pythagorean ($a,b\in K$ implies $\sqrt{a^2+b^2}\in K$), and that $K^2\cap\partial D=\emptyset$. Then $K^2\cap D$ is a semi-elliptic plane, and every semi-elliptic plane arises in this way. The metric is up to scale the restriction of the Klein-Beltramy metric on $D$ (logarithm of the cross-ratio), so the completion is the hyperbolic plane $D$ itself. Since according to Pejas's classification every Archimedean H-plane is Euclidean, hyperbolic or semi-elliptic (Theorem 3 in Greenberg's paper) the completion is always Euclidean or hyperbolic.

Since $K$ is only Pythagorean but not Euclidean (every positive element has a square root) there are lines in $K^2\cap D$ that pass through interior of circles without intersecting them in violation of the so-called line-circle axiom. This can also be seen geometrically. In any Archimedean H-plane that satisfies the hypothesis of the acute angle and the line-circle axiom a Bolyai's construction produces a line asymptotically parallel to any given one, which can not happen in $K^2\cap D$. The "metric constant" $k\in K$ is the Gaussian curvature of the completion, and since $\sqrt{-k}\notin K$ it can not always be normalized. In other words, not every hyperbolic plane can be thinned out into a semi-elliptic one, the standard one can't be among others. Pejas's example of $K$ is obtained by adjoining to $\mathbb{Q}(\sqrt{2})$ the square roots of all elements $x=a+b\sqrt{2}$ that are positive along with their conjugates $\overline{x}=a-b\sqrt{2}$, and taking a metric constant $k<0$ with $\overline{k}>0$, for instance $k=1-\sqrt{2}$. The equation of $D$ is then $x^2 +y^2<-k^{-1}$.

I'd like to thank Will Jagy for pointing me in the right direction in his previous answer