Pole/Zero plot is the same for passive and active (w/ gain) bandpass filter, why?
To compare how the system responded compared to a passive bandpass
  filter seen below, I used derived the transfer function and placed the
  same values used for the other filter, thus giving the same cut-off
  frequencies.


Wrong! You can't do this to compare your filters because now, there is an interaction between components that you would not get with a virtual-ground inverting op-amp configuration. It's even reflected in your bode plots - look at the mid-point frequencies - the op-amp circuit is about 10 kHz whilst the passive one is about 1 kHz. Look at the gains - they are massively different.

No, you can't do this and obtain anything useful to say.


  What do the poles and zeros represent in this system, and also
  shouldn't the opamp introduce its own poles and zeros and its is
  adding gain to the system.


A good designer will normally choose an op-amp so that its unwanted characteristics are avoided so no, most op-amp circuits do not rely significantly on the op-amps non-idealities.

Pole zero tutorial for a 2nd order system: -



The top three pictures are example bode plots of a 2nd order low pass filter. The bottom right is the traditional pole-zero diagram. The bottom left shows how the bode plot and pole-zero diagram are interlinked in a 3D way.