What is the use of resistors in series with BJTs in a push pull amplifier configuration?
You are correct that the circuit has a voltage gain of 1, and that the diodes are to minimize crossover distortion.  The idea is that the diode voltage is the same as the base-emitter voltage of the transistors, so that both transistors remain biased in the "on" condition - they are both conducting.  This means that there is always a theoretical current from the positive rail to the negative rail, through the pair of output transistors, which is wasted power and is dissipated as heat in the transistors and the series resistors. 

With no series resistors, all of this power must be dissipated in the transistors.  Consider what happens when the circuit starts to drive a load.  The transistors start to heat up, and the base-emitter drop starts to decrease to a value lower than in the relatively cool diodes. This causes the current through the two transistors to increase - further increasing heat by further decreasing the base-emitter drop and increasing the through-current.  So in general, make the series resistors as large as you can consistent with your output load characteristics.

As to the difference in circuit "A" and "C" from "B," think about the fact that the phase shift due to the capacitance of the diodes and transistors will not be identical.  This is not going to be noticed at low frequencies, but if you crank up the frequency on your simulation, you will see if the mismatch is pronounced enough to cause any issues.  If there is a problem, you can overcome some of the effect at high frequency from the diodes by using two large capacitors as shown in "B."