How to make MC34063A work as a step-up converter with 2.1V input?
Worst case, you cannot do it without 'cheating'.
 Vin_min is 1.8V and Vstart_TYPICAL is 2.1V.
 Using the ...E suffix part would be far better.

There are several ways of cheating.
One is to use the "Start" signal, whether push button or digital signal etc to boost a pre-charged capacitor to supply a higher initial voltage to allow starting. Once started the IC can then self power itself from a higher voltage. The switch element must of course still draw power from the 1.8V minimum battery, but that is not a problem.

A capacitor charged via  a Schottky diode will charge to within less than 0.2V of the charging supply. A Schottky diode will drop say 0.4V when discharging into modest load.  

So:  


Charge capacitor to 1.8-0.2 = 1.6V.
Step bottom of cap from ground to 1.8V with start signal.

Vcap is now 1.8+1.6 = 3.4V.  
Feed capout to IC VCC = 3.4 - 0.4 = 3V.  
IC will now start and Vcc can be bootstrapped.   

Cap has to be large enough to power circuit during startup phase.