Simple op-amp frequency response
For voltage gain G, isn't Ao = G?


There are two gains: The open loop gain and the closed loop gain. 

\$A_0\$ is the open loop gain a low frequencies (before the cutoff frequency) What you have called G is the closed loop gain, which is the -R2/R1 for the inverting configuration you have. 

With that, the open loop gain of the opamp over frequency could be modeled as:

$$ A_{ol}=\dfrac{A_0}{\frac{s}{\omega _b}+1}$$

Once you pass the cutoff frequency, the gain decays at a rate of 20dB/dec.

In reality, the closed loop gain is also frequency dependent (it has a bandwidth). If you start from the definition of the output voltage for an opamp: Vo=Aol(V+ - V-) you arrive at:

$$G=\dfrac{-R_2/R_1}{\frac{s}{\omega _A}+1} $$

Where \$\omega_A=\frac{\omega _t}{1+R_2/R_1} \$.

Hope this helps.

Add:

As per your new question in the comments—I think you are confusing 2 things. First, I don't know the naming convention your book uses, but when you say:


  
  wb = wt / (1 + R2/R1)
  wb = wt / Ao
  


There is something wrong about this. I mean, one is the cutoff frequency for the closed loop gain (1) and the other (2) is the cutoff frequency for the open loop one. Yet, they're both named the same in your text.

For the inverting amp, the closed loop gain is already given:

$$G=\dfrac{-R_2/R_1}{\frac{s}{\omega _A}+1}$$

For the noninverting one:

$$G=\dfrac{1+R_2/R_1}{\frac{s}{\omega _A}+1} $$

And the cutoff frequency, \$\omega_A\$, is the same for both (which is your first definition of \$\omega _b\$).
\$A_o\$ is indeed the open loop gain at low frequencies.