Can "Taxicab geometry" be given a Hilbert-style axiomatization? Hilbert's axioms provide a synthetic system for Euclidean geometry. Is it possible to do the same thing for the Taxicab plane? It would seem that one would only need to alter the axioms for congruence, since all the other properties are the same as in the Euclidean plane, and the congruence axioms are the ones that determine the metric properties of the plane. If so, how? Note that if we remove SAS and leave all the other axioms in place, Taxicab geometry becomes a model of the Hilbert axioms, but can we add some more axioms in place of it that make it the unique model? If so, which ones, and if not, why not?
 A: Note:
This is not even remotely a definitive answer to this question, so please don't accept it. What follows is too long for a comment, but may help with understanding the problem. (Also, it may not help with understanding the problem, which is another reason to not accept it. In any case, it would be better to leave the question open, i.e. with no accepted answer, in order to attract a definitive answer which could then be accepted. I am also very interested/invested in this question too and would like it to one day receive a definitive answer, which this answer most certainly is not. Since accepting this answer might discourage receiving a definitive one, please do not accept it.)
BEGINNING OF UNINTERESTING INTRO
I assume you already know most of what will be in this introduction, I am just stating it so as to organize my thoughts and to write as cogent of an answer as possible.

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*A system of axioms is "consistent" if there exists at least one model satisfying the axioms.

*A system of axioms is "complete" if there exists only one model (up to model isomorphism) satisfying the axioms.

*What we know already is that the axioms for Euclidean geometry are complete.

*What we also know is that the axioms for Euclidean geometry $-$ SAS are consistent, but not complete, because both taxicab and Euclidean geometry satisfy them.

*Thus our goal is to find a complete set of axioms such that taxicab geometry is the unique (up to isomorphism) model. Since the axioms for Euclidean geometry $-$ SAS are consistent, and taxicab geometry satisfies them, we can assume that this hypothetical complete set of axioms will contain the axioms for Euclidean geometry as a subset.

One thing to consider first would be the following question:

Are there other geometries, besides taxicab geometry, which satisfy all of Hilbert's axioms for Euclidean geometry except for perhaps the SAS postulate?
In other words, how "close" is this consistent set of axioms close to being complete?

If we can characterize all, or at least most, of such geometries which satisfy all of the axioms of Euclidean geometry except SAS, we will have a better idea of how to add additional axioms to restrict the resulting geometry to being taxicab geometry.
This is essentially the analog of the approach taken historically first with the parallel postulate -- we ask if there are other models of geometry which satisfy the axioms of neutral geometry (Euclidean minus parallel postulate), we find one, hyperbolic geometry.
Then we ask how close to being "complete" is neutral geometry -- well, it turns out, fairly complete, since hyperbolic geometry is actually a model of neutral geometry plus the negation of the parallel postulate, and not only that, but upon further investigation it turned out to be the only model of neutral geometry plus the negation of the parallel postulate.
Since the parallel postulate is either true or false (assuming the law of excluded middle), this shows that there are only two models of neutral geometry up to isomorphism.
END OF UNINTERESTING INTRO
Since removing the SAS postulate from Hilbert's axioms does not prevent any model of the resulting geometry from still being isomorphic (as a model) to the set $\mathbb{R}^2$ (with some metric), as the example of taxicab geometry itself shows,
A good place to look would be $\mathbb{R}^2$ with various metrics besides the Euclidean metric.
In particular, we know that the Euclidean axioms with SAS are complete, that taxicab geometry satisfies the negation of SAS, so really our question reduces to:

We know that (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) is consistent, since taxicab geometry satisfies it -- how close is it to being complete?

In particular, if taxicab geometry were the only model of this axiom system (up to isomorphism), we would be done. But if it is not the only model, then we would want to characterize the entire class of models satisfying the axiom system [(Euclidean geometry) $-$ (SAS) $+$ (negation of SAS)] as thoroughly as possible, because we could then basically just add on as axioms whatever makes taxicab geometry unique among this class, guaranteeing a complete axiom system

Our question has become: what makes taxicab geometry unique among the class of models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) ?

Now the reason why this answer is not a complete answer to your question is because I don't have an exhaustive characterization of the class of all models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) -- in fact, I only have what seems to be a possible subset, but I don't have any rigorous proof that in fact (1) these are all models for this axiom system, and (2) these are the only models for this axiom system. In fact, I only conjecture that (1) is true:

Conjecture: ($\mathbb{R}^2, d_p$), with $d_p$ the metric induced by the $p$-norm, is a model for (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) for every $p \not=2$.

Basis for the conjecture: As far as I can tell, the SAS postulate seems to be directly connected with isotropy of norms, i.e. whether or not the norms are "rotationally invariant", or whether or not there are a set of preferred directions, how strongly the norm is determined by its isometry group, how many "rotations" said isometry group has, or whether or not the norm is induced from an inner product.
A $p$-norm on $\mathbb{R}^2$ is isotropic, has no preferred directions, is "rotationally invariant", is completely determined up to scalar-multiple by its isometry group, and is induced by an inner product, if and only if $p = 2$.
You mentioned in the comments how the geodesics are not unique for the $L^1$ metric -- in fact, it seems to be the case that the geodesics are unique (and correspond exactly to straight lines, rather than just having them as a subset) for a metric induced by an $L^p$ norm if and only if $p = 2$. See this question. So uniqueness of geodesics may also be related to isotropy somehow.
See (1)(2)(3)(4)(5)(6)(7)(8) -- not all of these are directly relevant, but basically my point is that it is widely believed (rightly or wrongly) that having an inner product is necessary "to be able to talk about angles in a sensible way", and that my conjecture is that the SAS postulate is the synthetic geometry axiom which says that "we can talk about angles in a sensible way".
In other words, the negation of SAS may be equivalent to "metric is not induced by an inner product". It may be the case that (Euclidean geometry) $-$ (SAS) also somehow implies "space if "flat" (i.e. isomorphic to $\mathbb{R}^2$ as set), and the metric is induced by a norm".
If that is so, then we would even have that the class of all models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) is equivalent to "$\mathbb{R}^2$ with some norm besides the Euclidean norm". That would mean that the only additional axiom to add, besides (negation of SAS), would just be (norm is the $L^1$ norm) (although that raises the open question of how to phrase that in terms of synthetic geometry -- I don't know, I'm not even 100% certain how to phrase the existence of an inner product in terms of synthetic or metric geometry, much less a specific $p$-norm for $p \not=2$).
To summarize, there are several levels of uncertainty for me regarding this conjecture:

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*Are all of the proposed models ($\mathbb{R}^2$ with $p$-norm for $p \not = 2$) actually models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS)?


*What is the entire class of models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS)?


*Does every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) consist of $\mathbb{R}^2$ with some metric? Or do there exist models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) which are "not flat", i.e. not $\mathbb{R}^2$?
(At the very least, Hilbert's dimension axioms and second-order continuity schema should most likely ensure that any model is at the very least a 2-dimensional metrizable manifold, although I'm not even 100% certain of that. Still, I think we don't have to worry about things which look locally like $\mathbb{Q}^2$ or other oddities like that.)

*

*Assuming that every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is "flat" (and metrizable), do the axioms of (Euclidean geometry) $-$ SAS $+$ (negation of SAS) guarantee us at the very least that the metric is translation-invariant and homogeneous, i.e. induced by a norm?


*Assuming that every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with a norm-induced metric, does the norm have to be a $p$-norm?
Certainly, any norm which isn't the Euclidean norm isn't isotropic, not just $p$-norms for $p \not = 2$, which is why I only dare to conjecture that every $p$-norm for $p \not= 2$ satisfies (Euclidean geometry) $-$ SAS $+$ (negation of SAS), but that the converse is not true, i.e. that there may exist models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) which are not $\mathbb{R}^2$ with a $p$-norm for $p \not=2$.
Still, the latter might not matter if you can devise a synthetic geometry axiom which reduces to "the norm-induced metric is induced by a $p$-norm" and then a third synthetic geometry axiom "$p = 1$".
I don't know if the above actually makes any sense as written -- it makes sense in my head, but that's not a very good criterion for communicability.
To conclude:

(Really bold) Conjecture: Any and every model of (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with some translation-invariant and homogeneous but non-isotropic metric.
I.e., any and every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with any norm-induced metric (not even necessarily $p$-norm induced) besides the Euclidean metric.

If really bold conjecture is true, then as I tried to explain above, then theoretically at most two more synthetic axioms may be needed to add on to (Euclidean geometry) $-$ SAS $+$ (negation of SAS) in order to get a complete axiom system describing taxicab geometry.
