This is a different way to draw the proportionally-compensated closed-loop block diagram:
The assumption behind the closed-loop block diagrams you know (like the one shown in PetPaulsen's answer above) is that K1=K2=Kp If that's true you can simply replace K1 and K2 with a gain of Kp after the summer and you have the same mathematical relationship of inputs to ouptuts but just in the form you're used to.
So what happens when Kp=1? Since Kp=K1=K2 we already know that K1 and K2 are 1. The system is still compensated because the feedback gain (K2) is not zero. Thus, in this form it's apparent that when Kp=1 the system is still compensated and the controller will not be ignored. The above diagram shows what the canonical form can't show as clearly. In this example, it's easy to see that the system only collapses to an uncompensated system if K2=0 and K1=1, but not if they're both 1.
Another of your questions was basically 'Why is Kp=0 considered 'stable'? Shouldn't that be 'disconnected' instead?'
It's something of a controls convention that zero gain represents an uncompensated system. When you redraw the block diagram in the form I did you can begin to see the outlines of their thinking: if Kp=0, then K2=0 and feedback is broken and we're left with the uncompensated system. But isn't K1 supposed to equal K2, which is 0? Here's where a bit of fudging takes place (in my opinion). They like mathematical continua and zero almost fits perfectly: if Kp is the smallest positive number, or the smallest negative number then you have a proportional controller. In fact if it's anything other than 0 it works - just not quite zero. Unless you adjust your thinking: K1 only has to equal K2 when if you want a proportional feedback controller. If you break the feedback you're under no obligation to have K1=K2, so let's just make it 1 and call it a day.
That still doesn't make it practical though. Trust me, if the 'correct' answer to a problem is a gain of Kp=0 you're having your leg pulled.