Before I answer your question I would like to make this comment about the Liouville equation: it is a necessary condition on $\rho$, but it is not sufficient. This means that the correct $\rho$ must satisfy the Liouville equation, but not every $\rho$ that satisfies the Liouville equation is the correct probability density of microstate.
Turning to your first question, the answer is, no, $\rho(H)$ is not the only form that satisfies the Liouville equation. But the fact that $\rho(H)$ is an acceptable solution is important because it ties $\rho$ to a mechanical observable, the energy of the system. The premise of thermodynamics is that the macroscopic state is defined by a small number of variables, $(E,V,N)$, for example. A theory that would give $\rho=\rho(\Gamma)$ as a function of the microstate $\Gamma$ is of no use if $\Gamma$ is not something that we can observe and control experimentally.
Gibbs noticed that $\rho(H)$ is an acceptable solution and set out to guess mathematical forms of $\rho(H)$ that made thermodynamic sense. His first guess was the canonical distribution $\rho = c e^{-H/\Theta}$. Gibbs explained his reasons for choosing the exponential form as follows (Gibbs uses $P$ for what we have called $\rho$):
The distribution represented by
\begin{equation}\tag{90}
\eta=\log P = \frac{\psi-\epsilon}{\Theta},
\end{equation}
or
\begin{equation}\tag{91}
P = e^{\dfrac{\psi-\epsilon}{\Theta}},
\end{equation}
where $\Theta$ and $\psi$ are constants, and $\Theta$ positive, seems to represent the most simple case conceivable, since it has the property
that when the system consists of parts with separate energies,
the laws of the distribution in phase of the separate parts are
of the same nature, a property which enormously simplifies
the discussion, and is the foundation of extremely important
relations to thermodynamics.
"Seems to represent the most simple case conceivable" is a modest way of saying "a stroke of genius"!
