Your confusion is caused because "EMF" has, similarly to "voltage", somewhat different meanings in different contexts. Some sources define total EMF for a path as integral of total electric field over that (possibly closed as in your question above) path. This sounds reasonable as total electric field is usually what drives mobile charges in real conductors, but unfortunately it is not how induced EMF is defined in the context of AC circuits.
When we talk about induced EMF in AC circuits, we really mean integral of induced electric field only. Induced field, in AC circuits, is only a part of the total electric field; there is also the Coulomb field due to electric charges distributed on the conductor surface and sources.
In case of a perfect core-less inductor with terminals A,B the induced EMF is defined based on induced field $\mathbf E_i$ only:
$$
emf_{AB} = \int_A^B \mathbf E_i \cdot d\mathbf s.
$$
This can be expressed as
$$
emf_{AB} = - L\frac{dI}{dt}.
$$
One can't define induced EMF on the inductor based on total electric field and get this result. "Total field" EMF for a path is much lower than induced EMF for the same path, because the conductor counteracts the induced field with its own field. In a perfectly conducting wire segment, the Coulomb field and the induced field completely cancel each other.
The induced EMF can, however, be calculated using the Faraday law
$$
\oint_\gamma \mathbf E \cdot d\mathbf s = - \frac{d\Phi}{dt}
$$
where $\gamma$ is a closed curve, $\mathbf E$ is total electric field and $\Phi$ is magnetic flux through that closed curve.
This is accomplished by completing the path that goes inside the conductor from terminal A to terminal B by a path from B to A that does not follow the coils (it can be straight line from B to A or it can be curved somewhat in order to avoid the conductor but must not follow the coils closely).
The left-hand side is approximately equal to emf as defined, because we can re-express it using the induced field and the Coulomb field. Contribution from the Coulomb field to the integral is zero, so the Faraday law can be also stated using induced field only:
$$
\oint_\gamma \mathbf E_i \cdot d\mathbf s = - \frac{d\Phi}{dt}.
$$
Since contribution from the added path is negligible (the added path is short and not following the lines of induced electric field), this integral is approximately equal to emf as defined:
$$
emf_{AB} \approx \oint_\gamma \mathbf E_i \cdot d\mathbf s.
$$
The right hand side can be expressed in terms of magnetic field inside or current that flows in the inductor. Magnetic flux $\Phi$ is approximately equal to magnetic field inside the inductor $B$ times the area of one coil $A$ times the number of coils in the inductor $n$:
$$
\Phi \approx n A B.
$$
So we arrive at
$$
emf_{AB} \approx -nA\frac{dB}{dt}
$$
and this can be further re-expressed as $-L\frac{dI}{dt}$.
The case of a closed ring is different in that the Coulomb field, while it can be present, can't cancel the induced field everywhere in the conductor - the Coulomb field can't have non-zero circulation (line integral over closed loop). So total induced field itself has to be zero. How is that possible? Because even if external bodies produce their induced field $\mathbf E_{i,ext}$, the induced current distribution in the ring, howsoever weak, will produce counteracting induced electric field $\mathbf E_{i,ring}$ and these two cancel each other in the perfect conductor.
So in case of a ring, it doesn't matter whether we integrate the induced field or the total field over the closed loop. The result is the same - zero.
A ring made of perfect conductor will indeed have zero total electric field inside and thus both the first ("total field") and the second ("induced") kind of EMF is zero.
This does not mean electromagnetic induction does not happen, only that induced electric field due to external sources is exactly cancelled by induced electric field due to electric charges on the ring itself.