With the exception of you your claim (from the diagram) that the junction of the two diodes is 0V (it's not, it is +6V), I think your reasoning is sound.

In fact, what you have observed is a fundamental flaw in this design, that both transistors are in fact on, at least partially. That will naturally lead to a (fairly) short circuit between the two supply rails via the transistors, and this is indeed a problem. It's evident in a simulation:

^{simulate this circuit – Schematic created using CircuitLab}

I've reduced R1 and R2 to exaggerate the problem, but the point is that there's a large current flowing down through Q1 and Q2 even without any load connected. The usual solution is to add "ballast " (emitter degeneration) in the form of resistances between the emitters:

^{simulate this circuit}

Since each of these resistors R3 and R4 will develop a small voltage across them, due to quiescent current, they have the effect of reducing the potential difference \$V_{BE}\$ between the base and emitter of their respective transistor. Consequently, the transistors tend to switch off somewhat, reducing quiescent current.

Obviously this technique increases the output resistance of the system, which is not ideal, but as long as the load impedance is significantly larger in comparison, it's a small price to pay for reduced quiescent current.

Often we employ negative feedback in a closed loop to overcome this shortcoming. For illustration I'll replace the diodes with a fixed 1.4V source, because that's what R1, R2, D1 and D2 are there to do:

^{simulate this circuit}

I've added a heavy load, R5, and this is a plot of \$V_{IN}\$ (blue) and \$V_{OUT1}\$ (orange, left circuit open loop):

As you can see, the output is an attenuated version of the input, so gain is a little less than unity due to the emitter resistors.

On the right though, by closing the loop, the op-amp is able to compensate for the undesired attenuation. I won't plot it here, because the input and output waveforms are identical, but you can simulate it for yourself.

With the exception of your claim (from the diagram) that the junction of the two diodes is 0V (it's not, it is +6V), I think your reasoning is good.

In fact, what you have observed is a fundamental flaw in this design, that both transistors are indeed switched on, at least partially. That will naturally lead to a (fairly) short circuit between the two supply rails via the transistors. That's a big problem with this design, and is evident in a simulation:

^{simulate this circuit – Schematic created using CircuitLab}

I've reduced R1 and R2 to exaggerate the problem, but the point is that there's a large current flowing down through Q1 and Q2 even without any load connected. The usual solution is to add "ballast " (emitter degeneration) in the form of resistances between the emitters:

^{simulate this circuit}

Since each of these resistors R3 and R4 will develop a small voltage across them, due to quiescent current, they have the effect of reducing the potential difference \$V_{BE}\$ between the base and emitter of their respective transistor. Consequently, the transistors tend to switch off somewhat, reducing quiescent current.

Obviously this technique increases the output resistance of the system, which is not ideal, but as long as the load impedance is significantly larger in comparison, it's a small price to pay for reduced quiescent current.

Often we employ negative feedback in a closed loop to overcome this shortcoming. For illustration I'll replace the diodes with a fixed 1.4V source, because that's what R1, R2, D1 and D2 are there to do:

^{simulate this circuit}

I've added a heavy load, R5, and this is a plot of \$V_{IN}\$ (blue) and \$V_{OUT1}\$ (orange, left circuit open loop):

As you can see, the output is an attenuated version of the input, so gain is a little less than unity due to the emitter resistors.

On the right though, by closing the loop, the op-amp is able to compensate for the undesired attenuation. I won't plot it here, because the input and output waveforms are identical, but you can simulate it for yourself.