How do non-flowing common mode or bias currents cause common mode range to exceed/float?
But how come a 10k resistor tied between the node b and the AGND solve this issue?
  


Schematics are much better than words. Draw it out. 


  
  icm1 and icm2 has no return path, since the source is floating.
  


Correct. So there is no common mode current. 


  
  Most importantly how does icm1 and icm2 exist at all if there is no return path for these currents to loop?
  


It doesn't. 


  
  If there is no return path for icm1 and icm2 then how does it create a problematic voltage at node a and node b?
  


Because if Z &lt;&lt; R then \$ V_{ab} = V_{BAT} \$.









simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Instrumentation amplifier showing bias currents.

With no return path to the power supply the inputs will be driven to one supply rail (positive with the bias current direction shown) or the other. You can consider these as constant current generators - a bit like your lab bench supply on constant current mode. If there is no significant load the voltage will rise (or fall) to the limit of the power supply. 

Given that you have a voltage source between the two inputs it is likely that if the amp is sourcing current that the + input will rise to V+ and the - input will be held at \$ V_{batt} \$ below that.







simulate this circuit

Figure 2. A constant current power supply feeding a load, \$ R_L \$.

How it works:


R1 biases Q3 on. Current flows through R2, Q3 and RL.
When the current through R2 rises to about 10 mA there will be enough voltage drop to turn on Q4. This will steal the bias from Q3 and the circuit will settle with 10 mA through RL. The voltage at NODE1 will be 10m x RL.


Now what happens to the voltage at NODE1 if you disconnect RL? The current through R2 will decrease, Q4 will shut off, Q3 will turn hard on and NODE1 will rise to V+.

This is a very crude explanation and I have no idea what the input stage of your amplifier looks like. It might help.