EDIT, 7:51 pm: the argument below, except for the quadratic form and the entries in the matrices $B,C,$ is exactly the argument that says, once you know that the equator is a geodesic on the unit sphere, then all the geodesics on the sphere are the great circles, based on the fact that the great circles are the intersections of the sphere with planes through the origin.
ORIGINAL: We are looking at the hyperboloid sheet
$$ z = \sqrt{1 + x^2 + y^2 } $$
or
$$ x^2 + y^2 - z^2 = -1, \; \; z > 0.$$
This is in Minkowski $(2+1)$ space, with quadratic form $x^2 + y^2 - z^2$ as a semi-Riemannian metric. All orientation preserving isometries of the space are the product, in whatever order you like, of the three matrices below:
$$ A \; = \;
\left( \begin{array}{ccc}
\cos \theta & \sin \theta & 0 \\
- \sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}
\right) ,
$$
$$ B \; = \;
\left( \begin{array}{ccc}
\cosh u & 0 & \sinh u \\
0 & 1 & 0 \\
\sinh u & 0 & \cosh u
\end{array}
\right) ,
$$
$$ C \; = \;
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & \cosh v & \sinh v \\
0 & \sinh v & \cosh v
\end{array}
\right) ,
$$
Of course, these also preserve the sheet $ x^2 + y^2 - z^2 = -1, \; \; z > 0,$ and the subspace metric on the sheet (positive definite, needs separate proof) is preserved by definition.
Because the curve $$ (\sinh t, \; 0, \; \cosh t )$$ is invariant pointwise under the orientation reversing isometry
$ y \rightarrow -y,$ it is a geodesic curve that lies in a plane $y=0$ through the origin. It can thus be carried to any other geodesic curve by the isometries. However, the isometries not only fix the origin, they take planes through the origin to to other planes through the origin.
That's about it.