Why doesn't the particle in a ring model have a potential energy term?
The short answer is that the particle in a ring is defined to have no potential energy term (at least, not beyond what Karsten Theis has pointed out). Once you introduce something else into the system, then it is no longer really the particle on a ring problem any more.

However, if you want to relate it to the hydrogen atom, then note that in the hydrogen atom, the Coulomb potential is a function of only $r$, i.e.

$$V(r) = -\frac{1}{r}$$

in atomic units. In the particle on a ring model system, the particle is already constrained to be at a particular value of $r$, so even if we were to hypothetically introduce a Coulomb-type potential $\propto 1/r$, this would just be a constant. The effect of having a constant potential energy is just to shift every eigenstate up in energy by the same amount; it has no real physical effect.

What are the differences in the mathematical solutions? Well, you already know it, surely; the eigenstates of the hydrogen atom are atomic orbitals, whereas the eigenstates of the particle on a ring are simply $\exp(\mathrm im\phi)$ with $m \in \mathbb{Z}$ to satisfy the boundary condition.

Of course, this $\exp(\mathrm im\phi)$ term is indeed part of the mathematical form of the atomic orbitals. This is because when you solve the Schrödinger equation for the hydrogen atom, you can successively separate out different degrees of freedom. Firstly you separate the bit which depends on $r$, which goes on to become the radial wavefunction $R(r)$. Then you need to solve for the angular wavefunction $Y(\theta,\phi)$, and the way of doing this is to again use separation of variables, i.e. assume $Y(\theta,\phi) = f(\theta)g(\phi)$. That gets you to a differential equation which looks like $(-1/g)(\mathrm d^2g/\mathrm d \phi^2) = m^2$, which has $g = \exp(\mathrm im\phi)$ as its solutions. I'm glossing over some details here, but that's the general idea; consult e.g. Griffiths' Introduction to Quantum Mechanics for the nitty-gritty bits.