Let $T_{20}$ be the initial temperature of tank 2 and $T_{10}=T_{20}+\Delta T$ be the initial temperature in tank 1. Let $\delta T$ be the equal rise in the temperature of both thanks. Assuming that the piston does not move, we would have $$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$
Since $p_1=p_2$, if we divide one equation by the other, we obtain:
$$\frac{p_{2f}}{p_{1f}}=\frac{(T_{20}+\delta T)(T_{20}+\Delta T)}{T_{20}(T_{20}+\Delta T+\delta T)}=1+\frac{(\Delta T)( \delta T)}{T_{20}^2+T_{20}(\Delta T+\delta T)}$$
So, if the piston doesn't move, the final pressure in chamber 2 will be higher than in chamber 1. The piston must move in the direction from chamber 2 to chamber 1.
I quote from the abstract of one of the articles which the OP linked in the question preceding this one
The principle of increase of entropy is used to obtain the condition for mechanical equilibrium in an isolated system divided into two parts by a frictionless, weightless piston which is made of a perfectly thermally insulating material. The result emphasizes that the principle can be used to obtain the condition for mechanical equilibrium without the assumption, frequently made in the textbooks, that the mechanical equilibrium is accompanied by thermal equilibrium.
It seems to be explicitly stated here that Callen's "flaw" is this assumption. See also this physics.SE question for an explanation of the difference between mechanical and thermal equilibrium in one of the answers.
To summarise in my own words: $P_1 = P_2$ only makes sure that you are in a state where nothing is moving/accelerating (i.e. that you are in mechanical equilibrium), which would obviously be the case if $P_1 \neq P_2$ (thus it is a necessary condition, as the OP states). However this is not sufficient to conclude that you are in thermodynamic equilibrium, because due to thermal fluctuations you always get $P_1 \neq P_2$ some of the time. Now it could be the case that the $(P_1 = P_2)$-state was metastable and drifts away from the mechanical equilibrium induced by the small fluctuations. As far as I can tell from the articles linked in the other question that is not the case, so $P_1 = P_2$ does turn out to be the correct condition, but Callen did not rigorously show that the "restoring force" accompanying the fluctuations leads back to the mechanical equilibrium.
I am not 100% sure if this answers the question, nor if it is correct. In particular I personally think that Callen's argument can easily be made complete by noting that the mechanical equilibrium $P_1 = P_2$ is unique. I mean if your system leaves mechanical equilibrium due to fluctuations it has to go somewhere. And there is no second mechanical equilibrium for it to go to. I.e. necessary $\Leftrightarrow$ sufficient since it is also unique.
Update
This is to address the update to the question (v6).
Additionally to the original question the update contains an outline of the argument that is supposed to be correct, yielding the same outcome as Callen's argument. The OP then points out that the arguments seem to be the same.
I don't think this is the case. As stated previously by the OP Callen only uses the first law while for a correct proof of the equilibrium the second law should be employed. The reason for this is what I tried to point out above in my original answer. The first law is a mere statement of energy conservation. You can't extract information about thermodynamic equilibrium by simply imposing the constraints of your system. That only shows that mechanical equilibrium is realised, which is a necessary but not sufficient condition for thermodynamic equilibrium (as the OP already stated repeatedly).
The second law provides the sufficiency. Why? Because it is a statement about the nature of thermodynamic equilibrium itself, in particular it tells you how fluctuations behave around the point of equilibrium and that you will bounce back into your original state under fluctuation perturbations, to say it from the statistical mechanics viewpoint.
One of the OPs doubts is
Indeed, from (4) and (9) we would obtain $dU_2 = -P_2 dV_2$ (analogos to (1)): this, together with (1), (2) and (10), would give us Callen's argument again!
Of course your are going to get the same relations as from the first law if you plug the relations obtained from the second law argument into each other, because mechanical equilibrium happens to be thermal equilibrium in the situation considered. However I fail to see how this coincidence means that they are the same argument, since they start from completely different points.
Best Answer
You've discovered a famous problem in thermodynamics.
In our case the piston will not move. The mechanical argument is right, while the maximum entropy argument is inconclusive.
To see that $P_1=P_2$ is an equilibrium position you can also apply conservation of energy. Since there is no heat exchange,
$$dU_{1,2} = -P_{1,2} dV_{1,2}$$
We require that $dU=0$ since our system is isolated from the environment, hence
$$dU_1 + dU_2 = 0 \to P_1 d V_1 + P_2 dV_2 = 0$$
But $V=V_1+V_2$ and $V$ is fixed, so that $dV_1 = - dV_2$ and we obtain
$$P_1=P_2$$
Now let's see the entropy maximum principle. The problem is that you forgot that $S$ is a function of energy too:
$$S(U,V)= S_1 (U_1, V_1)+ S_2 (U_2, V_2)$$
$$d S = dS_1 + dS_2 = \frac{dU_1}{T_1} + \frac{P_1}{T_1} dV_1 + \frac{dU_2}{T_2} + \frac{P_2}{T_2} dV_2$$
Since $dU_{1,2} = -P_{1,2} dV_{1,2}$, we see that $dS$ vanishes identically, so that we can say nothing about $P_{1,2}$ and $T_{1,2}$: the entropy maximum principle is thus inconclusive.
Update
Your question actually inspired me a lot of thoughts in the past days and I found out that...I was wrong. I basically followed the argument given by Callen in his book Thermodynamics (Appendix C), but it looks like:
My error was really silly: I only showed that $P_1=P_2$ is a necessary condition for equilibrium, not that it is a sufficient condition, i.e. (if the argument is correct and) if the system is at equilibrium, then $P_1=P_2$, but if $P_1=P_2$ the system could still be out of equilibrium...which it is!
I am still not really able to explain why the whole argument is wrong: some authors have said that equilibrium considerations should follow from the second law and not from the first and that the second law is not inconclusive. You can read for example this article and this article. They both use only thermodynamics considerations, but I warn you that the second tries to contradict the first. So the problem, from a purely thermodynamic point of view, is really difficult to solve without making mistakes, and I have found no argument that persuaded me completely and for good.
This article takes into consideration exactly your problem and shows that the piston will move, making the additional assumption that the gases are ideal gases.
This is really similar to the argument given by Feynman in his lectures (39-4). Feynman basically uses kinetic theory arguments to show that if we start with $P_1 \neq P_2$ the piston will at first "slosh back and forth" (cit.) until $P_1 = P_2$, and then, due to random pressure fluctuations, slowly converge towards thermodynamic equilibrium ($T_1=T_2$).
The argument is really tricky because we assume that if the pressure is the same on both side the piston will not move, forgetting that pressure is just $2/3$ of the density times the average kinetic energy per particle
$$P = \frac 2 3 \rho \langle \epsilon_K \rangle$$
just like temperature is basically the average kinetic energy (without the density multiplicative factor). So we are dealing with statistical quantities, which are not "constant" from a microscopic point of view. So while thermodynamically we say that if $P_1=P_2$ the piston won't move, from a microscopic point of view it will actually slightly jiggle back and forth because of the different collision it experiences from particles in the left and right sides.
There have been also simulations of your problem which show that if we start with $P_1=P_2$ and $T_1\neq T_2$ the piston will oscillate until we reach thermodynamic equilibrium ($T_1=T_2$). See the pictures below, which I took from the article.