Why a pure resistive AC circuit absorbs power at all times according to this equation
You haven't quoted the two special cases of Eq. (11.10) so we can't comment on that.

I suspect that what the author is suggesting is that if the equations demonstrate that the maximum possible power is being extracted from the AC then the circuit is purely resistive.
Since \$ V_{rms} = \frac {1}{\sqrt 2} V_{max} \$ and \$ I_{rms} = \frac {1}{\sqrt 2} I_{max} \$ then \$ P =  V_{rms}I_{rms} = \frac {1}{\sqrt 2} V_{max} \frac {1}{\sqrt 2} I_{max} = \frac {1}{2} V_{max}I_{max} \$. A reactive circuit would always give a lesser value for power.


  I understand that resistor only absorb power and there is no reactive elements in the circuit like capacitors and inductors ...


Be careful with that thought. The circuit can contain inductors and capacitors and appear to be purely resistive if their reactances are equal and opposite. This is the principle of operation of industrial power-factor correction where capacitors are switched in across the mains supply to counter the inductance of (typically) the motors used in the plant. Correcting the power factor means that the electricity network current is kept to a minimum and efficiency is kept high.

Note that a power-factor correction scheme such as discussed above will be balanced at a certain frequency but that's not a problem on the electrical grid as it runs at constant frequency.