In the case that $H$ is trivial, $\omega$ is just a $\frak{g}$-valued coframing on the manifold $P$. The curvature $2$-form $\Omega = d\omega + \tfrac{1}{2}[\omega,\omega]$ can then be written in the form
$$
\Omega = R(\omega\wedge\omega)
$$
where $R:P\to {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)$ is the curvature tensor. Since $H$ is trivial, there is no group action to take into account, so, technically, any $\lambda\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ could be used to construct a scalar-valued curvature quantity $\lambda(R):P\to\mathbb{R}$.

Now, you seem to want to consider the scalar curvatures constructed from those $\lambda$ that are invariant under the adjoint action of $G$ on $\frak{g}$, and typically there are such elements. For example, the Lie algebra structure itself is an element $\mu\in {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)=\mathrm{Hom}\bigl(\Lambda^2({\frak{g}}),{\frak{g}}\bigr)$. In the semi-simple case, one can use the Killing form to dualize this to an element $\mu^\ast\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, and then the resulting 'scalar curvature' $\mu^\ast(R)$ is a very natural invariant.

More generally, though, you might want to ask for the $G$-invariant elements in ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ and see if there is anything else interesting there. However, in the case that $\frak{g}$ is simple, it is not hard to show that the induced action of $G$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ only fixes multiples of $\mu^\ast$.

*Remark:* One possible reason for considering the $G$-invariant elements is this: There is a natural action of $G$ on Cartan connections for the pair $(G,\lbrace e\rbrace)$, namely, one can define $a\cdot \omega = \mathrm{Ad}(a)\bigl(\omega\bigr)$. Then one will have $\lambda\bigl(R(\omega)\bigr)=\lambda\bigl(R(a\cdot\omega)\bigr)$ when $\lambda$ has this kind of $G$-invariance.

The more general case, when $H$ is nontrivial, can be handled similarly, and you'll want to look at the action of $H$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g/h}})$ and consider the invariant elements $\lambda$. [NB: I have now modified this sentence to take into account the OP's comment below, as he pointed out that he wasn't restricting to torsion-free geometries.]

As to which of these invariants might be thought of as comparing some kind of local volume with some kind of 'geodesically parallel' volume (which is what the scalar curvature does in Riemannian geometry), I'd have to think about this some more.

*Example:* In the case that $G$ is the group of rigid motions of $\mathbb{R}^n$ and $H\simeq\mathrm{SO}(n)$ is the group of rotations about a point (in which the OP was particularly interested), one has
$$
{\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n
\qquad\text{and}\qquad
{\frak{h}}\simeq {\frak{so}}(n),
$$
and the usual Cartan connection on the orthonormal frame bundle $P\to M$ of a Riemannian metric $g$ on an $n$-manifold $M$ is of the form
$$
\omega = (\theta,\eta):TP\to {\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n
$$
where $\eta$ is the 'soldering form' and $\theta$ is the Levi-Civita connection form. The curvature of $\omega$ as a Cartan connection is then
$$
\Omega = \bigl(d\theta+\theta\wedge\theta,\ d\eta + \theta\wedge\eta\bigr)
= \bigl(R(\eta\wedge\eta),0\bigr),
$$
where, now, $R:P\to \mathrm{Hom}\bigl(\Lambda^2(\mathbb{R}^n),{\frak{so}}(n)\bigr)
\simeq {\frak{so}}(n)\otimes {\frak{so}}(n)$ is the usual Riemann curvature tensor.
In this case, because the Cartan connection is torsion-free, the Cartan connection curvature takes values in
$$
{\frak{h}}\otimes\Lambda^2\bigl(({\frak{g}/\frak{h}})^\ast\bigr)\simeq
{\frak{h}}\otimes\Lambda^2\bigl(\mathbb{R}^n\bigr) \simeq
{\frak{so}}(n)\otimes {\frak{so}}(n)
$$
and, applying the above recipe to this curvature, the unique $H$-invariant element (up to multiples) in this space does indeed give the usual scalar curvature of the underlying Riemannian metric (up to a constant multiple).

NB: When $n=4$, the above space ${\frak{so}}(n)\otimes {\frak{so}}(n)$ actually has a $2$-dimensional space of vectors fixed under the action of $H$, but, of course, the first Bianchi identity says that the curvature takes values in a proper subspace of ${\frak{so}}(n)\otimes {\frak{so}}(n)$ that only intersects this $2$-dimensional space in a $1$-dimensional space after all. That brings up the point that, if one does consider only Cartan connections that satisfy some given torsion restrictions, then there can be Bianchi identities that say that the curvature of the connection must take values in some proper subspace of ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, in which case, one will have to take this into account in considering which 'scalar curvatures' can be defined.