In general, the notion of isometries for Riemannian manifolds is an infinitesimal one. A smooth map $\varphi \colon (M,g) \rightarrow (N,h)$ between smooth Riemannian manifolds is called a local isometry if for each $p \in M$ the differential $d\varphi_p \colon (T_pM, g_p) \rightarrow (T_{\varphi(p)}N, h_{\varphi(p)})$ is an isometry of inner-product spaces. A local isometry is a local diffeomorphism - if it is also a global diffeomorphism then it is called an isometry.
Given this definition, the answer to your question is yes - local isometries preserve the Riemann curvature tensor and since the curvature tensor for $\mathbb{R}^n$ is identically zero the curvature tensor of $M$ vanishes identically.
Note that the relevant term is "the Riemann curvatures" and not "Gaussian curvature" as the Gaussian curvature is usually a scalar that is associated to a surface embedded in $\mathbb{R}^m$ and you are asking about a general $n$-dimensional Riemannian manifold.
Regarding your other questions - local isometries of Riemannian manifolds preserve the length of paths and preserve the distance induced by the Riemannian metrics (the infimum of the length of all paths connecting two points). The definition of a local isometry presented above requires the map $\varphi$ to be at least differentiable. You can also define an isometry as a map that preserves the distances in the sense of metric spaces. Such a map will be continuous but a priori doesn't have to be differentiable. It turns outs that a map that preserves the distances in the sense of metric spaces will automatically be an isometry in the sense defined above and in particular, smooth. This result is called Myers-Steenrod theorem.
To show that straight lines from the origin are geodeics, you can make use of the fact that isometries preserve geodesics. More precisely, if $\psi:M\to N$ is an isometry between Riemannian manifolds ($M$ and $N$ may be the same) and $\gamma:(a,b)\to M$ is a path in $M$, then $\gamma$ is a geodesic in $M$ if and only if $\psi\circ\gamma$ is a geodesic in $N$. A proof of this will depend on what definition of a geodesic you're using.
Additionally, we know from ODE theorems on existence and uniqueness of solutions that if $\gamma(0)$ and $\dot{\gamma}(0)$ are specified, there is a unique maximal geodesic $\gamma:(a,b)\to M$ satisfying these initial conditions.
Combining these two, it's straightforward to show that if $\psi:M\to M$ is an isometry, $\gamma:(a,b)\to M$ is a maximal geodesic, $\psi(\gamma(0))=\gamma(0)$, and $d\psi(\dot{\gamma}(0))=\dot{\gamma(0)}$, then $\psi\circ\gamma=\gamma$. That is, if the position and velocity of a geodesic is invariant under an isometry at one point, then the entire geodesic is invariant under that isometry.
Using the above claim, we can make a symmetry argument. In the Hyperbolic metric on the $2$-ball we can choose a geodesic $\gamma$ with $\gamma(0)=(0,0)$ and $\dot{\gamma}(0)=(1,0)$. Since these are invariant under the isometry $\psi(x,y)=(x,-y)$ we conclude that $\psi\circ\gamma=\gamma$, and thus $\gamma$ must lie entirely on the $x$-axis. This determines $\gamma$ up to parameterization. Since rotations about the origin are also isometries, all of the rotations of $\gamma$ are also geodesics.
As for the point about distance functions, it's important to remember that the Riemannian metric is an inner product on tangent spaces, not on the manifold itself. The construction you describe would give a distance function on a tangent space, but not on the underlying manifold. That said, it is possible to derive a distance function on a connected Riemannian manifold, namely
$$
d(p,q)=\inf\left\{\text{length}(\gamma):\gamma\text{ a piecewise differentiable curve joining $p$ and $q$}\right\}
$$
With this distance function, a connected Riemannian manifold is a metric space, and the two different notions of isometry always agree. Establishing this requires a bit more work than on vector spaces, though, and it's not particularly useful for this problem.
Best Answer
If the Riemannian metric you are using here is the restriction of the usual scalar product $\langle\cdot,\cdot\rangle$ of $\mathbb{R}^{n+1}$, that is $$\langle\cdot,\cdot\rangle_p=\langle\cdot,\cdot\rangle|_{T_pS\times T_pS},$$ then as soon as $T_pS=T_qS$, you will have $\langle\cdot,\cdot\rangle_p=\langle\cdot,\cdot\rangle_q$. This is indeed the case here since $T_pS=p^\perp=(-p)^\perp=T_{-p}S$. If the Riemannian metric depends of the point, then it can no longer be true.