I'm not sure exactly what you mean by 'absolute geometry' that unifies spherical, Euclidean, and hyperbolic geometry, but in the sense of Klein's identification of geometries with group actions, so that an 'underlying' geometry is an extension of a group action, there is a very specific answer to your question.
There are two distinct geometries that 'underlie' each of the spherical, Euclidean, and hyperbolic geometries in dimension $n$: conformal geometry and projective geometry. Moreover, there is no finite dimensional Lie transformation group that extends either of these geometries.
In the case of conformal geometry, the simply-connected model space is $\mathbb{S}^n$, regarded as the quotient of $\mathrm{O}^+(n{+}1,1)$ by the subgroup that fixes a future-directed null ray in the Lorentzian space $\mathbb{R}^{n{+}1,1}$, i.e., $\mathbb{S}^n$ is the space of future-directed null rays in $\mathbb{R}^{n{+}1,1}$. To specialize to the spherical case, one takes the subgroup $G_+\subset\mathrm{O}^+(n{+}1,1)$ that fixes a future-directed timelike vector $v_+\in\mathbb{R}^{n{+}1,1}$. Then $G_+$ acts transitively on $\mathbb{S}^n$ and preserves a metric of positive sectional curvature $+1$ on $\mathbb{S}^n$. To specialize to the Euclidean case, one takes the subgroup $G_0\subset\mathrm{O}^+(n{+}1,1)$ that fixes a future-directed null vector $v_0\in\mathbb{R}^{n{+}1,1}$ and notes that $G_0$ acts transitively on the set $\mathbb{S}^n\setminus \{\mathbb{R}^+v_0\}$, preserving a (flat) Euclidean metric. To specialize the hyperbolic case, one takes the subgroup $G_-\subset\mathrm{O}^+(n{+}1,1)$ that fixes a spacelike vector $v_-\in\mathbb{R}^{n{+}1,1}$ and notes that $G_-$ acts transitively on the hemisphere $\mathbb{H}^n = \{ \mathbb{R}^+v\in \mathbb{S}^n | \langle v_-,v\rangle <0 \} $ preserving a Riemannian metric of sectional curvature $-1$.
In the case of projective geometry, the simply-connected model space is $\widetilde{\mathbb{RP}^n}$, the space of rays in $\mathbb{R}^{n+1}$, regarded in a natural way as the quotient of $\mathrm{PGL}(n{+}1,\mathbb{R})$ by the subgroup that fixes a specific ray $r_0\in \mathbb{R}^{n+1}$. Spherical geometry is recovered by restricting to the subgroup $\mathrm{PO}(n{+}1)\subset \mathrm{PGL}(n{+}1,\mathbb{R})$, which acts transitively on $\widetilde{\mathbb{RP}^n}$ preserving a Riemannian metric with sectional curvature $+1$, and hyperbolic geometry is recovered by restricting to the subgroup $\mathrm{PO}(n,1)\subset \mathrm{PGL}(n{+}1,\mathbb{R})$, which acts transitively on a hemisphere of $\widetilde{\mathbb{RP}^n}$ preserving a Riemannian metric with sectional curvature $-1$, etc.
Note that, while $\mathbb{S}^n$ and $\widetilde{\mathbb{RP}^n}$ are diffeomorphic, the two groups $\mathrm{O}^+(n{+}1,1)$ and $\mathrm{PGL}(n{+}1,\mathbb{R})$ are not isomorphic when $n>1$. They don't even have the same dimension. Hence, these two 'underlying geometries' are not equivalent in any sense. Moreover, it is not hard to show that the only Lie transformation group on $\mathbb{S}^n$ that properly contains $\mathrm{O}^+(n{+}1,1)$ is the full group of diffeomorphisms of $\mathbb{S}^n$ and that the only Lie transformation group on $\widetilde{\mathbb{RP}^n}$ that properly contains $\mathrm{PGL}(n{+}1,\mathbb{R})$ is the full group of diffeomorphisms of $\widetilde{\mathbb{RP}^n}$. Thus, these two incompatible underlying geometries are, in a sense, 'minimal' nontrivial underlying geometries.