What happens when transistor operates in saturation region for constant current source
when the forward voltage of the LED is increased from 1.7V to 3.3V,
  and our Emitter voltage is maintained at 2V...


In saturation &beta; reduces towards zero as VCE goes to zero, so the Base draws more current. In your circuit that will cause greater voltage drop across RB1, so the Emitter voltage won't be maintained at 2V. Instead it will go down, causing the Collector current to also go down. The LED will get 3.3V, but at lower current.  

Here's the result of simulating your circuit in LTspice, with LED voltage varying from 1.7V to 4.5V:-



Between 1.7V and 2.9V the transistor does a good job of keeping the LED current (red line) constant. 

Above 2.9V the the transistor goes into saturation as the voltage between the  Collector (green line) and Emitter (blue line) drops below 0.2V, causing &beta; to reduce and requiring more Base current (magenta line) to maintain current through RE. But the higher Base current also reduces Base voltage as it draws more current through RB1, which in turn reduces the Emitter voltage. With less voltage across RE (as well as more current coming from the Base instead of from the Collector) the Collector current also reduces. In this region the transistor is acting more like a resistor than a constant current source.

If the voltage divider was 'stiffer' and held the Base voltage constant despite the increased Base current then the transistor would go into hard saturation, with the Base supplying enough Emitter current to keep VE close to 2V even if the Collector current dropped to zero. With a 5V supply and the Emitter at 1.8V there would not be sufficient voltage left to light a 3.3V LED.