Model of ordered plane which is neither isomorphic to $\mathbb{R}^2$ nor to Klein model Let $B_{\mathbb{R}}\subset\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ be standard (strict) betweenness relation on $\mathbb{R}$ i.e.
$$B_{\mathbb{R}}(abc):\iff\left(a<b<c \vee c<b<a\right)$$
My definition of ordered plane is as follows:
$P$ is a set (plane), $\mathcal{L}\subset 2^P$ is a family of lines and $B\subset P\times P\times P$ is a ternary betweenness relation. We say that $(P,\mathcal{L},B)$ is an ordered plane whenever


*

*$(P,\mathcal{L})$ is a model of Hilbert's incidence axioms.

*$B(abc)$ implies that $a,b,c$ are collinear.

*For any line $L$: $(L,B|_{L\times L\times L})$ is isomorphic to $(\mathbb{R},B_{\mathbb{R}})$.

*Pasch's axioms holds.


I know two standard models of this axioms i.e. $\mathbb{R}^2$ and Klein model and I'm looking for a model which is not isomorphic to any of these two. 
Basically these axioms are a variant of Hilbert's neutral geometry axioms excluding congruence relations. Actually I think it's a bit more because every line is isomorphic to $\mathbb{R}$ and maybe there are models of Hilbert's axioms without congruence (but with continuity) in which at least one line is not isomorphic to $\mathbb{R}$ (This is something I am asking as well).
It is known that neutral geometry has exactly two models up to isomorphism but excluding congruence should make it have more models. Note that these axioms involve (Dedekind) continuity axiom and as a result there are no countable models such as "rational" plane.
 A: Here are two examples.
The Moulton plane
As suggested in the comments. This is a classical construction which keeps all the points of the Euclidean plane, but modifies the lines slightly. Vertical lines, and lines $y = mx + b$ with slope $m \ge 0$, stay the same. Instead of lines with negative slope, we take the sets of points satisfying $$y = \begin{cases}mx + b & x < 0, \\ 2mx + b & x \ge 0\end{cases}$$ for some $m,b \in \mathbb R$ with $m<0$. That is, lines with negative slope "refract" as they cross the $y$-axis.
We can check that this satisfies all the incidence axioms and Pasch's axiom, and that each line is still isomorphic to $\mathbb R$ as far as betweenness is concerned. The interesting bit is proving that the Moulton plane is not just a weird model of the Euclidean or the hyperbolic plane, which we could fix by defining congruence in a slightly different way.
The distinction is that the Moulton plane violates Desargues's theorem. Essentially, you can draw a configuration in the Moulton plane where Desargues's theorem is violated by first drawing it in the Euclidean plane, then positioning it so that exactly one of the lines (say, the axis of perspectivity) has negative slope, and bends away before it touches one of the points it's supposed to go through.
Meanwhile, both the Euclidean plane and the hyperbolic plane satisfy the theorem, which is a way to distinguish them from the Moulton plane without using congruence. (The Moulton plane also satisfies the parallel postulate, which distinguishes it from the hyperbolic plane).
The subset $(0,1) \times \mathbb R$
This one has the advantage of being simple to understand. You take the subset $(0,1) \times \mathbb R$ of the points of the Euclidean plane, and have it inherit all the lines of the Euclidean plane that pass through it. Vertical lines are obviously isomorphic to $\mathbb R$, because they haven't lost anything; all other lines look like open intervals, which are also isomorphic to $\mathbb R$ as far as betweenness is concerned.
We distinguish this model from both Euclidean and hyperbolic geometry via the parallel postulate - say, in the form of Playfair's axiom. Unlike Euclidean geometry (where the axiom always holds) and hyperbolic geometry (where it is always violated), this model distinguishes between vertical lines and all other lines.
Given a vertical line $\ell$ and a point $P$ not on $\ell$, there is only one line through $P$ parallel to $\ell$: the vertical line passing through $\ell$.
Given a horizontal or diagonal line $\ell$ and a point $P$ not on $\ell$, we can draw infinitely many lines through $P$ parallel to (that is, not intersecting) $\ell$. 
