Ultimately the explanation, like you said, requires us to invoke special relativity. However, this explanation might be opaque at first, so I'll add some physics-y interpretation after sketching the SR calculation.
The electric and magnetic fields form components of a 2-dimensional tensor, $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, where $A_\mu$ is a four-vector$^\star$ whose components are given by the normal scalar and vector potentials: $A_0=\phi$ and $A_i$ are the components of the vector potential $\vec{A}$, in units where $c=1$.
The reason to bring this up is that it doesn't make sense to say you only have a magnetic field in two different frames. In one of the two frames you mentioned, you must have an electric field. This is a consequence of the transformation laws
\begin{eqnarray}
\vec{E}'_{||} &=& \vec{E}_{||} \\
\vec{B}'_{||} &=& \vec{B}_{||} \\
\vec{E}'_{\perp} &=& \gamma (\vec{E}_{\perp} + \vec{v} \times \vec{B}_\perp )\\
\vec{B}'_{||} &=& \gamma (\vec{B}_{\perp} - \vec{v} \times \vec{E}_\perp )
\end{eqnarray}
where primed quantities are evaluated in the boosted frame and unprimed in the original frame, $\gamma=\sqrt{1-v^2}$ is the usual Lorentz factor, $\vec{B}_{||}$ refers to the components of $\vec{B}$ parallel to the velocity of the boost, $\vec{B}_\perp$ refers to the component of the magnetic field perpendicular to the velocity, I've made the same definitions for the electric field, and as a reminder I've set $c=1$.
So now we apply these formulas. Let's suppose that in the original frame, the ring is moving in the $x$ direction, and the field is purely magnetic and the magnetic field is pointed in the $z$ direction. Then the Lorentz force $\vec{F}\propto \vec{v}\times \vec{B}$ will point in the $y$ direction. Then in the rest frame of the ring, we find that there must be an electric field with a $y$ component given by
\begin{equation}
E_y' = - \gamma v_x B_z
\end{equation}
This electric field provides the force in the $y$ direction you were worried about.
OK but this doesn't exactly answer the question of where does this electric force come from? The answer is that you have made an idealization by considering an infinitely large magnetic field, and in doing so you've lost track of what is generating the magnetic field in the first place. There must evidently be some currents "at infinity", but no net charge because there is no electric field. You need to account for length contraction when you apply the boost, which will tend to create a net charge density, following an analysis present in (for example) the book by Purcell.
Having said all of this, what will happen in this case is that negative charges will tend to "pile up" at the maximum and "deplete" at the minimum of the ring along the $y$ direction. This charge separation will provide a counter electric force that will oppose and eventually stop any current. So there is not a current when the ring reaches equilibrium. This may be a reason why this case is not typically studied explicitly.
Einstein did consider a related problem as motivation for special relativity. He not consider the problem of a constant magnetic field over all space, but rather a finite conductor generating the magnetic field which could generate a steady current via induction. The fact that in different frames there were different explanations for why there was a current (motional emf vs the Lorentz force) bothered him, and he states his motivation for developing special relativity in his classic 1905 paper is to give a deeper explanation that works in any inertial reference frame.
$^\star$ well, if you are careful about gauge invariance, $A_\mu$ isn't really a 4 vector, but let's ignore this.
