The simple answer is: Don't risk it and get ones that are rated for inductive loads. Here's the relevant Wikipedia article on power factor. Basically you have electrical impedance which consists of electrical resistance (which is opposition to movement of current created by resistors) and reactance (which is opposition to movement of current by all kinds of coils and by capacitors). In DC systems, the reactance is zero.
Actual reactance depends very much on the circuit itself and it can be zero and then we say that the circuit is resistive, it can be greater than zero and we say that the circuit is inductive and if it's lower than zero, we say that circuit is capacitive. By adding capacitors to inductive circuit we can make it less inductive, resistive and, if we add enough capacitors, capacitive. Same goes for other way around. If we add enough coils to capacitive circuit in the end we can get inductive circuit.
The problem here is that it's not uncommon to find device that doesn't provide enough data to easily determine if it's capacitive or inductive, what its power factor is and it can be difficult even when all that is known to calculate if the total load on the circuit breaker is resistive, capacitive or inductive. Another point is that the circuit-breakers linked don't provide (or at least I can't find) enough information to determine when an inductive load is too inductive for them.
So to be safe, just get circuit-breakers that can break an inductive load.
The basic reason for using a parallel tuned circuit as the exciter is one of efficiency. If your inductor and tuned capacitor are 33uH and 75nF, resonant frequency will be about 101kHz. If you do the math you'll see that a lossless tuned circuit like this exhibits infinite impedance but still circulates high current between cap and inductor.
Lossless circuit are of course impossible but making the losses as low as possible means that if your inductance is 33uH and your applied voltage is (say) 40VRMS at 100kHz, the current in the inductor is: -
\$\dfrac{40V}{2\cdot\pi\cdot F\cdot L}\$ = 1.93 Amps
Your H bridge won't even be breaking into a sweat because it won't be supplying anything like this current. This current is due to the voltage across the inductor but the capacitor has the effect of performing "power-factor" correction and because the losses are low maybe the H bridge will be supplying in the order of 50mA to a couple of hundred mA.
However, your H bridge is exciting the coil/cap with a square wave and there will be losses due to the harmonics within the square wave. Because of this it makes sense to feed the coil/cap via an inductor too - somewhat smaller than the coil (maybe 1 quarter). You will also need to retune the capacitance to compensate for this. Some experimentation in this is required to get best results but, you should aim to reduce the H bridge's current to avoid it overheating.
I'd also say make one larger coil suitable for all three inductive loads. The larger coil can be any regular shape that suits your requirements for placement of the receiving coils.
Optimum performance is when the receiving coils are also tuned with a capacitor but, because the induced voltage is in series with a receive coil, the tuned circuit behaves like a series tuned circuit and, if the coupling is too great it will heavily detune the transmit coil when it is close by. You should aim for a minimum gap or incorporate circuits in the H bridge that current limit.
I strongly advise you to use something like LTSpice for simulating this - you'll learn a lot about the various interactions. I'd also recommend you read a bit about tesla coils because that is what are are intending to build (when tuned as per my thoughts).
Best Answer
If the device contained nothing it would obviously be fraud, so they put in an inexpensive part. It is still fraud, but only informed people will know.