Doesn't Joule's Law of heating suggest that all the potential energy supplied is converted to heat in a resistor I know it is a repeated question, but the other questions couldn't clear my doubts. So, sorry.
In the Joule's Law of heating, we say $P=VI$ where V is the potential difference across the resistor and I the current through the resistor. In a simple circuit with one resistor and a battery, V is equal to the potential difference supplied by the battery. So now when we plug in the values of V and I, doesn't it mean that all of the potential energy is lost as heat. But if that is the case then how do they get to the other side of the circuit, since all energy was just lost as heat.
Another question is that, isn't it actually the kinetic energy of the electron that is lost and not the potential energy of electron.
Or is the reason for why we say its potential energy that is lost as follows: When electron loses its kinetic energy as heat that much energy leaves the system. So it is equivalent of saying that that much potential energy of the system was lost even before the electron has entered the resistor.
Please consider criticising my explanation too.
Thank you
 A: 
But if that is the case then how do they get to the other side of the
  circuit, since all energy was just lost as heat.

The two are not incompatible.  Imagine sliding a mass down a slope, and we have selected it so that the force of gravity and friction balance exactly.  Since there is no net force, the block does not accelerate.  The PE it loses by dropping down the slope goes into heat (friction losses).  
The same thing happens here.  Without resistance, the charge would be accelerated from the high potential point to the low potential point.  But because of the resistor, the energy is converted into heat instead of KE.  
When the circuit was first closed, there was a short time when some of the energy went into accelerating the charges to the steady-state current.  But after that time the energy only went into heating.  

Another question is that, isn't it actually the kinetic energy of the
  electron that is lost and not the potential energy of electron.

No, because the KE isn't changing.  Again, think of the block sliding down the slope at constant speed.  The net work done on the mass is zero, so no change in KE.  In the circuit, the change in potential and the resistance losses sum to zero, so no change in KE.

So it is equivalent of saying that that much potential energy of the
  system was lost even before the electron has entered the resistor.

It's not lost before the resistor, it's lost in the resistor.  The change in PE is exactly balanced by the resistance losses.  If it weren't, the charges in that section would speed up until the losses balanced.
