I can complete Anton's plan with an additional assumption that geodesics do not branch. I also assume local compactness (otherwise there are too many technical details to deal with). More precisely, I prove the following:
Let $X$ be a geodesic space. Suppose that
For every finite subset $Q\subset X$, every similarity $Q\to X$ can be extended to a bijective similarity $X\to X$. A similarity is a map that multiplies all distances by a constant. (It is easy to see that, in the case of a length metric, the more complicated definition from the question reduces to this.)
$X$ is locally compact.
For every two distinct points of $X$ there is a unique line containing them. (A line is a subset isometric to $\mathbb R$. The existence of lines follows from the two other assumptions.)
Then $X$ is isometric to $\mathbb R^n$ for some $n$.
Proof. First observe that the group of similarities acts transitively on pairs (oriented line, point not on this line). Indeed, given two such pairs $(\ell_1,p_1)$ and $(\ell_2,p_2)$, it suffices to find similar triples $p_1,x_1,y_1$ and $p_2,x_2,y_2$ where $x_i,y_i$ is a positively oriented pair of points on $\ell_i$. And it is easy to see that such triples $p_i,x_i,y_i$ realise all similarity types of, e.g. isosceles triangles.
Next we construct perpendiculars. Lines $\alpha$ and $\beta$ intersecting at a point $q$ are said to be perpendicular if there exists an isometry that fixes $\alpha$ and maps $\beta$ to itself by reflection in $q$. It is easy to see that this relation is symmetric. Further, for every line $\ell$ and every point $p\notin\ell$ there exists a unique perpendicular to $\ell$ containing $p$. To prove existence, pick two points $x,y\in\ell$ such that $|px|=|py|$. There is an isometry that fixes $p$ and exchanges $x$ and $y$. It acts on $\ell$ by reflection in the midpoint $q$ of $[xy]$, hence the line $(pq)$ is perpendicular to $\ell$. Uniqueness follows from the fact that two distinct reflections of $\ell$ cannot fix $p$, otherwise their composition shifts $\ell$ along itself and fixes $p$, and some iteration of this shift would break the triangle inequality.
This argument also shows that the base $q$ of the perpendicular is the nearest point to $p$ on $\ell$, and the distance from $p$ to $x\in\ell$ grows monotonically with $|qx|$.
Clearly all right-angled triangles with given leg lengths $a$ and $b$ are isometric; denote their hypotenuse by $f(a,b)$. Then $f$ is strictly monotone in each argument and positively homogeneous: $f(ta,tb)=tf(a,b)$.
Next, we show that the sum of Busemann function of two opposite rays is zero. Or, equivalently, if $\gamma$ is an arc-length parametrized line, $q=\gamma(0)$ and a line $(pq)$ is perpendicular to $\gamma$, then $B_\gamma(p)=0$ where $B_\gamma$ denotes the Busemann function of $\gamma$. Suppose the contrary. We may assume that $|pq|=1$ and $B_\gamma(p)=c>0$. This means that for every point $x\in\gamma$ we have $|px|-|qx|\ge c$. Let $s\in\ell$ be very far away and let $p_1$ be the perpendicular from $q$ to $(ps)$. The above inequality implies that $|pp_1|\ge c$, hence we have an upper bound $|qp_1|\le\lambda<1$ where $\lambda$ is determined by $f$ and $c$ and does not depend on $s$. Let $q_1$ be the base of the perpendicular from $p_1$ to $(qs)$, then rescaling the above inequality yields that $|p_1q_1|\le\lambda|qp_1|\le\lambda^2$. The next perpendicular (from $q_1$ to $p_2$ on $(ps)$) has length at most $\lambda^3$ and so on. Summing up these perpendiculars, we see that $|ps|\le 1/(1-\lambda)$ which is not that far away, a contradiction.
The fact that opposite rays yield opposite Busemann functions implies that the Busemann function of a line $\gamma$ is the only 1-Lipschitz function $f$ such that $f(\gamma(t))=-t$ for all $t$. And, given local compactness and non-branching of geodesics, this implies that $X$ is split into lines parallel to $\ell$, where a line $\gamma_1$ is said to be parallel to $\gamma$ if the Busemann function of $\ell$ decays with unit rate along $\gamma_1$ (to construct a parallel line, just glue together two opposite asymptotic rays). Further, Busemann functions of parallel lines coincide, due to their above mentioned uniqueness.
Now it is easy to see that the level set of a Busemann function (which is also a union of perpendiculars to a given line at a given point) has the same isometry extension property and is a geodesic space. Then we can carry induction in the parameter $d$ defined as the maximum number of pairwise perpendicular lines that can go through one point. The cases $d=1$ and $d=2$ can be done by hand, then use isometries exchanging perpendicular lines to do the induction step.
Here is an expanded version of my previous comments. There are a lot of things to check here and I haven't.
In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. Tarski works in a system where the domain is $\mathbb{R}^2$ and you have two relations, a ternary betweenness relation and an equidistance relation $E(x,y;x',y')$ which says that the line segments $\overline{xy}$ and $\overline{x',y'}$ are congruent. In this system all the axioms are phrased in terms of points, not in terms of both points and lines like in Euclid. Lines emerge as definable sets.
The Tarski axioms for Euclidean geometry are on Wikipedia, and you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,×,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof.
But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane.
Edit: I think that my initial attempt at the Hyperbolic model of the Euclidean plane fails, because equidistance isn't definable. Following a suggestion of @ColinMcLarty, I will describe a different model that I think works. This model is from Greenberg's book "Euclidean and Non-euclidean geometries".
Let $\mathbb{H}$ be the Hyperbolic plane. In this model the points are just the points in $\mathbb{H}$ and the lines all all lines in $\mathbb{H}$ through the origin together with the curves in $\mathbb{H}$ that are equidistant from a line through the origin.
To get a model of Euclidean geometry we need a betweenness and equidistance relationship, and these relationships need to be definable in $\mathbb{H}$. The betweenness relationship is easy, you just need to observe that the lines form a uniformly definable family of sets. Equidistance is more complicated.
Greenberg describes a map $\rho : \mathbb{H} \to \mathbb{R}^2$ and says that the equidistance relation on $\mathbb{H}$ is the pull-back of the equidistance relation on $\mathbb{R}^2$ by $\rho$. We can put polar coordinates on $\mathbb{H}$ in the same way as on the Euclidean plane. We fix a ray $\ell$ through the origin and let the polar coordinates of $p \in \mathbb{H}$ be $(r,\theta)$ where $r$ is the Hyperbolic distance from the origin to $p$ and $\theta$ is the angle that $p$ makes with $\ell$. If $p \in \mathbb{H}$ is the point with polar coordinates $(r,\theta)$ then $\rho(p)$ is the point
$$
\rho(p) = (\sinh r \sin \theta, \sinh r \cos \theta) = \sinh r (\cos \theta, \sin \theta).
$$
So $\rho(p)$ is the point in $\mathbb{R}^2$ whose (euclidean) polar coordinates are $(\sinh r, \theta)$.
Now let's suppose that $\mathbb{H}$ is the Poincare disc model. I think it is enough to show that $\rho$ is semialgebraic, i.e. definable in $(\mathbb{R},+,\times,<)$. The Hyperbolic distance $r$ of $p$ from the origin is not a semialgebraic function of $p$, but it is the arcsinh of a semialgebraic functions, see Wikipedia. (On Wikipedia there is a factor of $2$ in from the arcsinh, but that is just a scaling factor, so we can drop it). I think it should be pretty clear that $\rho$ is semialgebraic.
Once we know that $\rho$ is semialgebraic we know that the equidistance relation on our Hyperbolic model of Euclidean geometry is semialgebraic. This should imply that it is definable in Hyperbolic geometry, but one would need extra arguments to see it directly.
Best Answer
It is a famous problem (due to Banach and Mazur) whether a separable infinite dimensional Banach space which has a transitive isometry group must be isometrically isomorphic to a Hilbert space. Of course, if every two dimensional subspace has a transitive isometry group, then the space is a Hilbert space since then the norm satisfies the parallelogram identity. For counterexamples in the non separable setting, consider the $\ell_p$ sum of uncountably many copies of $L_p(0,1)$ with $p$ not $2$.
For a recent paper related to the Banach-Mazur rotation problem, which contains some other references related to the problem, see
http://arxiv.org/abs/math/0110202.