Why is this derivation of Schmitt trigger circuit's output signal, incorrect?
You made an assumption there's a stable output and calculated what the output should be in case the assumption is right. In addition you clipped the result to the possible output voltage range. The clipping is ok, but the assumption of the existence of stable output is not, as persons who know the feedback stability theory could confirm (see NOTE1)
People make same kind of reasoning continuously. Actually the whole physics is based on this kind of reasoning. There comparing to measurements is the way to reveal wrong assumptions.
NOTE1: there's no need to be an academic level mathematician or engineer to be able to make it clear a stable output can be achieved only because the limited voltage range clips the output. Some elementary Laplace domain analysis is enough.
If we assume there's some realistic slowness in the amp, say one RC charging and the gain is finite, maybe big, but finite we can find the transfer function for the whole circuit. Th slowness prevents infinitely fast changes so we can follow what the circuit does.
We can replace the ideal amplification A with G/(1+sRC) which is the transfer function of buffered RC integrator. G is the DC gain of the amp.
Let's also simplify the formula by replacing R1/(R1+R2) with a single symbol B. It's our feedback attenuation factor which is between 0 and 1.
The s-domain gain of the system is Vo/Vi = 1/(B-(1+sRC)/G)
Of course the output stays zero if the input is zero and there's no noise. But there's always some noise. We can find which s-domain frequencies start to ring in the system from the slightest pulse of noise by calculating which values of s make the denominator infinite (=find the poles of the transfer function). We solve s from equation (B-(1+sRC)/G)=0
The result is s=(GB-1)/RC
The Laplace transformation math says that the output from a slightest pulse of noise is proportional to an exponential voltage exp(t/T) with time constant T=RC/(GB-1). This T is positive as soon as GB is bigger than 1. Positive time constant means limitless grow which is in practice stopped only by the limited output voltage range. Negative T (i.e. GB &lt;1) means that the ringing in the loop decays and the output stabilizes to the value which can be calculated with your original formula for Vo. But for a stable output A must be less than how much the feedback voltage divider attenuates.