How can I protect a MOSFET bi-directional switch
Should I connect the common? My guess is yes because in case of a
  event


You don't need to connect to common because one of the MOSFETs will be reversed biased and its body diode will be therefore forward biased.


  Because isolator is slow (say ~1mS), I am concerning about the moment
  when SW just opens or closes into inductive load (a motor)


Well, when activating, one of the MOSFETs is going to be vulnerable and the other will conduct through its body diode and be less of a problem but you do need to check that the body diode can handle the same current as the load when passing through the off-to-on situation.

By the looks of it the IRF540 is going to be OK in that respect.

But generally, and putting numbers on it, if the load is 20 amps and this activates in 1 ms linearly, the voltage will fall from (say) 50 volts to 0 volts and the peak power will be about halfway i.e. 10 amps and 25 volts = 250 watts. You could use a simulator to get a more precise figure (and I recommend that you do) but, a short estimate is 250 watts for 1 ms.

Looking at figure 8 in the data sheet suggests that it will be OK at these levels but you have to put real numbers onto this graph that represent the true operating conditions (and not just my guesstimates): -




  The plan is to put some uni-directional TVS across mosfets with VBR &lt;
  VDSS


The TVS would need to be quite beafy in order to handle the back emf at the current being taken by the inductive load when it is disconnected. This might be 10 amps (for instance) and the energy stored in the inductive load might keep pushing that current into the TVS for a few tens of milliseconds. If the TVS voltage was 70 volts, that's a power of 700 watts sustained for several tens of milli seconds.

I'd probably consider one of these: -



It has a stand-off voltage of 60 volts at 2 uA (only you can decide if this is sufficient) and it clamps at 96.8 volts with a peak current of 52.7 amps. If you draw a line between 60 volts and 96.8 volts and moved along that line a linear proportion of the current you will need (possibly 10 amps) compared to 52.7 amps, you can estimate that at 10 amps, the TVS will clamp around 67 volts.

Again, these are my numbers and you need to use your own. Anyway, that's a peak power of 670 watts and well within the 5 kW rating but the next graph will help you understand how long it will survive: -



As you should be able to see, a 10 ms pulse of nearly 2 kW is on the limit so, providing your pulse isn't much in excess of 10 ms you should be OK. But you can parallel devices providing you used worst case limits but I would recommend using a simulator for this. I designed a triple-parallel TVS surge suppressor like this.

Do not use MOV devices as an alternative - the MOV will be called into action for each on-off cycle of the load and MOVs/varistors only have a limited number of surge cycles before they fail short-circuit. A TVS (of the appropriate voltage rating) is, more than likely, the most reliable solution but choose with care and choose a beafy one.