Your logic is sound. More opamps means more gain for a given bandwidth.
Compensated op-amps are made with a single dominant pole so the gain/bandwidth product is constant. If this isn't working for your application, use a decompensated amp and do the compensation network yourself.
There is a simple answer: The bandwidth for the closed-loop gain is determined by the frequency where the LOOP GAIN is 0 dB. In your example circuits the loop gain is not the same - hence, the bandwidth will not be the same. The circuit with the largest loop gain (non-inverter) has the largest bandwidth.
Explanation why the Loop Gain (LG) determines bandwidth:
The denominator of the closed-loop gain formula is
\$ D(s) = 1 - LG \$
From this, we can derive that "something" happens when \$LG=1\$ (0 dB).
At the corresponding frequency \$ \omega_{o} \$ we have a real pole (think of the behaviour of a first-order lowpass). And this pole gives the frequency where the 3dB-bandwidth is defined.
I should add that this is a simplified explanation; a detailed explanation involves the open-loop gain Aol and its frequency response:
\$ A_{CL} = \dfrac{H_{FW} \cdot A_{OL}}{1 - Hr \cdot A_{OL}} \$
with \$LG=Hr * A_{OL}\$ and forward factor \$H_{FW}\$.
We can see that for low frequencies (large \$LG\$) and negative feedback factor (\$Hr\$ negative) the "1" can be neglected and the gain is
\$A_{CL}= \dfrac{H_{FW}}{Hr} \$ = constant.
However, for large frequencies (\$A_{OL}\$ and \$LG\$ smaller) we cannot neglect the "1". When we reach the frequency \$ \omega_{o} \$ where \$ |LG|=1\$ the "1" starts to dominate for larger frequencies and we can neglect the loop gain LG.
In this case the numerator \$H_{FW} \cdot A_{OL}\$ determines mostly the frequency response (\$ A_{CL}= H_{FW} \cdot A_{OL}\$, approximately a first order lowpass).
Hence, the transition from the first region to the second region is at the cut-off frequency wo.
For inverter: \$H_{FW}=\dfrac{-R2}{R1+R2}\$
For non-inverter: \$H_{FW}=1\$.
Best Answer
The point of characterizing the gain-bandwidth product is that you will find that no matter what gain you configure using reasonable feedback resistors you will find
$$\left| A\right|\cdot f_{-3dB} \approx GBW$$
where \$GBW\$ is a characteristic of a particular op-amp.
As said above, it doesn't matter what gain configuration you use, you will get pretty close to the same gain-bandwidth product.
Using the open-loop gain might be convenient. But using a gain of 10 or gain of -2 config might be more convenient.
Datasheets will normally say what gain they use, and it's not always open-loop, since loading effects and non-linearities will change the \$GBW\$ slightly.
You'll have to say exactly what formula you mean.
In the formula I gave above, \$A\$ is the linear gain, not the decibel gain. (So you'd use 100,000 rather than 100, if you were using the open-loop gain of an op-amp with a gain of 100,000 or 100 dB)