The issue is one of terminology, and this particular issue is an extremely common one. (In the context of classical thermodynamics, almost all of the issues that students run into are ones of terminology.)

In this case, you are confusing the notions of reversible/irreversible and quasi-static/non-quasi-static. Quasi-static is a term that applies to an individual system, and reversible is a term that applies to a collection of systems. A system undergoes a quasi-static (or quasi-equilibrium) process when it moves through a sequence of equilibrium states. A quasi-static process can be irreversible, if the irreversibility occurs in some sense outside the system; in the context of elementary thermodynamics, irreversibility is usually a consequence of heat flow across a finite temperature difference between a system and a reservoir (or, as in the OP's case, between one system and another).

Therefore, the short answer is that the OP went wrong when saying that $dS \neq \delta Q/T$ for the process. By assumption, the individual processes undergone by systems A and B are both quasi-static, and so $$dS_A = \frac{\delta Q_A}{T_A},$$ and $$dS_B = \frac{\delta Q_B}{T_B}.$$

So here is how I would analyze the problem, based on what the OP has done. Suppose we have the same setup, in which both systems A and B are at finitely different temperatures $T_A$ and $T_B$ to begin with, and they are then brought into contact via a rigid, impermeable, immovable, conducting boundary. In this case, the work done on either gas must be zero, as the OP notes, due to the membrane being rigid and immovable. In addition, since the processes undergone are quasi-static, we can write that the change in entropy of each subsystem during a small chunk of the process is $$dS = \frac{\delta Q}{T},$$ and that the change in internal energy during that small chunk of the process is $$dU = \delta Q + \delta W = TdS.$$ By conservation of energy, the total change in energy of the combined system is given by $$dU = dU_A +dU_B = \delta Q_A + \delta Q_B=0,$$ since the systems are in thermal contact but are otherwise isolated from the rest of the universe.

Now, we do know that this process must be irreversible, since there is heat flow across a finite temperature difference, so let's compute the total change in entropy of the combined system. Without loss of generality, let's take $T_A > T_B$, in which case $$\delta Q_B = -\delta Q_A = |\delta Q_A|.$$ (That is, $\delta Q_B>0$ since heat must flow into system B, and $\delta Q_A<0$ since heat must flow out of system A.) Then, the combined change in entropy during the process is \begin{align*} dS &= dS_A + dS_B = \frac{\delta Q_A}{T_A} + \frac{\delta Q_B}{T_B}\\ & = \frac{-|\delta Q_A|}{T_A} + \frac{|\delta Q_A|}{T_B} = |\delta Q_A|\frac{T_A-T_B}{T_AT_B} >0, \end{align*} where the last inequality arises as a consequence of $T_A > T_B$.

The issue is one of terminology, and this particular issue is an extremely common one. (In the context of classical thermodynamics, almost all of the issues that students run into are ones of terminology.)

In this case, you are confusing the notions of reversible/irreversible and quasi-static/non-quasi-static. Quasi-static is a term that applies to an individual system, and reversible is a term that applies to a collection of systems. A system undergoes a quasi-static (or quasi-equilibrium) process when it moves through a sequence of equilibrium states. A quasi-static process can be irreversible, if the irreversibility occurs in some sense outside the system; in the context of elementary thermodynamics, irreversibility is usually a consequence of heat flow across a finite temperature difference between a system and a reservoir (or, as in the OP's case, between one system and another).

Therefore, the short answer is that the OP went wrong when saying that $dS \neq \delta Q/T$ for the process. By assumption, the individual processes undergone by systems A and B are both quasi-static, and so $$dS_A = \frac{\delta Q_A}{T_A},$$ and $$dS_B = \frac{\delta Q_B}{T_B}.$$

So here is how I would analyze the problem, based on what the OP has done. Suppose we have the same setup, in which both systems A and B are at finitely different temperatures $T_A$ and $T_B$ to begin with, and they are then brought into contact via a rigid, impermeable, immovable, conducting boundary. In this case, the work done on either gas must be zero, as the OP notes, due to the membrane being rigid and immovable. In addition, since the processes undergone are quasi-static, we can write that the change in entropy of each subsystem during a small chunk of the process is $$dS = \frac{\delta Q}{T},$$ and that the change in internal energy during that small chunk of the process is $$dU = \delta Q + \delta W = TdS.$$ By conservation of energy, the total change in energy of the combined system is given by $$dU = dU_A +dU_B = \delta Q_A + \delta Q_B=0,$$ since the systems are in thermal contact but are otherwise isolated from the rest of the universe.

Now, we do know that this process must be irreversible, since there is heat flow across a finite temperature difference, so let's compute the total change in entropy of the combined system and see if it's strictly positive. Without loss of generality, let's take $T_A > T_B$, in which case $$\delta Q_B = -\delta Q_A = |\delta Q_A|.$$ (That is, $\delta Q_B>0$ since heat must flow into system B, and $\delta Q_A<0$ since heat must flow out of system A.) Then, the combined change in entropy during one tiny chunk of the process (so that the temperatures can be treated as constant) is \begin{align*} dS &= dS_A + dS_B = \frac{\delta Q_A}{T_A} + \frac{\delta Q_B}{T_B}\\ & = \frac{-|\delta Q_A|}{T_A} + \frac{|\delta Q_A|}{T_B} = |\delta Q_A|\frac{T_A-T_B}{T_AT_B} >0, \end{align*} where the last inequality arises as a consequence of $T_A > T_B$. This shows that at every step along the way, the entropy increases until the systems have equilibrated.

To complete the calculation, we merely integrate each expression separately. Assuming that each system has the same mass and that the specific heats are temperature-independent, we have \begin{align*} \Delta S_A &= \int \frac{\delta Q_A}{T_A}\\ &=\int_{T_A}^{(T_A+T_B)/2} \frac{n C dT}{T_A}= nC\ln\left(\frac{(T_A+T_B)/2}{T_A}\right), \end{align*} where $n$ is the number of moles and $C$ is the molar specific heat. The same expression holds for B (with B switched with A), and when we add the two expressions, it simplifies to $$\Delta S = \Delta S_A + \Delta S_B = 2 n C\ln\left(\frac{(T_A+T_B)/2}{\sqrt{T_AT_B}}\right).$$

The issue is one of terminology, and this particular issue is an extremely common one. (In the context of classical thermodynamics, almost all of the issues that students run into are ones of terminology.)

In this case, you are confusing the notions of reversible/irreversible and quasi-static/non-quasi-static. Quasi-static is a term that applies to an individual system, and reversible is a term that applies to a collection of systems. A system undergoes a quasi-static (or quasi-equilibrium) process when it moves through a sequence of equilibrium states. A quasi-static process can be irreversible, if the irreversibility occurs in some sense outside the system; in the context of elementary thermodynamics, irreversibility is usually a consequence of heat flow across a finite temperature difference between a system and a reservoir (or, as in the OP's case, between one system and another).

So here is how I would analyze the problem, based on what the OP has done. Suppose we have the same setup, in which both systems A and B are at finitely different temperatures $T_A$ and $T_B$ to begin with, and they are then brought into contact via a rigid, impermeable, immovable, conducting boundary. In this case, the work done on either gas must be zero, as the OP notes, due to the membrane being rigid and immovable. In addition, since the processes undergone are quasi-static, we can write that the change in entropy of each subsystem during a small chunk of the process is $$dS = \frac{\delta Q}{T},$$ and that the change in internal energy during that small chunk of the process is $$dU = \delta Q + \delta W = TdS.$$ By conservation of energy, the total change in energy of the combined system is given by $$dU = dU_A +dU_B = \delta Q_A + \delta Q_B=0,$$ since the systems are in thermal contact but are otherwise isolated from the rest of the universe.

Now, we do know that this process must be irreversible, since there is heat flow across a finite temperature difference, so let's compute the total change in entropy of the combined system. Without loss of generality, let's take $T_A > T_B$, in which case $$\delta Q_B = -\delta Q_A = |\delta Q_A|.$$ (That is, $\delta Q_B>0$ since heat must flow into system B, and $\delta Q_A<0$ since heat must flow out of system A.) Then, the combined change in entropy during the process is \begin{align*} dS &= dS_A + dS_B = \frac{\delta Q_A}{T_A} + \frac{\delta Q_B}{T_B}\\ = \frac{-|\delta Q_A|}{T_A} + \frac{|\delta Q_A|}{T_B} = |\delta Q_A|\frac{T_A-T_B}{T_AT_B} >0, \end{align*} where the last inequality arises as a consequence of $T_A > T_B$.

So really, the only part where the OP went wrong was in saying that $dS \neq \delta Q/T$ for this process. This is why getting the differences between quasi-static and reversible straight is so important.

The issue is one of terminology, and this particular issue is an extremely common one. (In the context of classical thermodynamics, almost all of the issues that students run into are ones of terminology.)

In this case, you are confusing the notions of reversible/irreversible and quasi-static/non-quasi-static. Quasi-static is a term that applies to an individual system, and reversible is a term that applies to a collection of systems. A system undergoes a quasi-static (or quasi-equilibrium) process when it moves through a sequence of equilibrium states. A quasi-static process can be irreversible, if the irreversibility occurs in some sense outside the system; in the context of elementary thermodynamics, irreversibility is usually a consequence of heat flow across a finite temperature difference between a system and a reservoir (or, as in the OP's case, between one system and another).

Therefore, the short answer is that the OP went wrong when saying that $dS \neq \delta Q/T$ for the process. By assumption, the individual processes undergone by systems A and B are both quasi-static, and so $$dS_A = \frac{\delta Q_A}{T_A},$$ and $$dS_B = \frac{\delta Q_B}{T_B}.$$

So here is how I would analyze the problem, based on what the OP has done. Suppose we have the same setup, in which both systems A and B are at finitely different temperatures $T_A$ and $T_B$ to begin with, and they are then brought into contact via a rigid, impermeable, immovable, conducting boundary. In this case, the work done on either gas must be zero, as the OP notes, due to the membrane being rigid and immovable. In addition, since the processes undergone are quasi-static, we can write that the change in entropy of each subsystem during a small chunk of the process is $$dS = \frac{\delta Q}{T},$$ and that the change in internal energy during that small chunk of the process is $$dU = \delta Q + \delta W = TdS.$$ By conservation of energy, the total change in energy of the combined system is given by $$dU = dU_A +dU_B = \delta Q_A + \delta Q_B=0,$$ since the systems are in thermal contact but are otherwise isolated from the rest of the universe.

Now, we do know that this process must be irreversible, since there is heat flow across a finite temperature difference, so let's compute the total change in entropy of the combined system. Without loss of generality, let's take $T_A > T_B$, in which case $$\delta Q_B = -\delta Q_A = |\delta Q_A|.$$ (That is, $\delta Q_B>0$ since heat must flow into system B, and $\delta Q_A<0$ since heat must flow out of system A.) Then, the combined change in entropy during the process is \begin{align*} dS &= dS_A + dS_B = \frac{\delta Q_A}{T_A} + \frac{\delta Q_B}{T_B}\\ & = \frac{-|\delta Q_A|}{T_A} + \frac{|\delta Q_A|}{T_B} = |\delta Q_A|\frac{T_A-T_B}{T_AT_B} >0, \end{align*} where the last inequality arises as a consequence of $T_A > T_B$.