Suppose a container separating two gases via a movable massless piston. If the pressure of gas 1 is higher than that of gas 2, the piston will move. In an infinitesimal volume change, the energy lost by gas 1 will be P1.dV and that gained by gas 2 will be P2.dV. As P1 is higher than P2, where is the extra energy going?
Note: no heat exchange is considered. The only change in energy is caused by work.
Let the system $1$ be one gas, the system $2$ be the other gas, and $P$ be the piston. The process is short enough to consider it adiabatic, so the first principle of thermodynamics gives
$$ \textrm{d}E_p + \textrm{d}U_1 + \textrm{d}U_2 = \delta W_{ext} $$
where $E_p$ is the mechanical energy of the piston and $W_{ext}$ the work done by the operator. Indeed, $\textrm{d}U_p = 0$ since the piston is a rigid body and there is no heat transfer, and $\textrm{d}E_1 = \textrm{d}E_2 = 0$ in the first order since the transformation is infinitesimal.
However, the process can be quasi-static, so $\textrm{d}E_p = 0$, so we got
$$ \textrm{d}U_1 + \textrm{d}U_2 = \delta W_{ext}$$
Here, you made the following mistake: you said that $\delta W_{ext} = 0$, but this is physically impossible. Indeed, $P_1 \neq P_2$, so the piston won't evolve smoothly at all if there's no external force applied on it. Thus, if the process is quasi-static, then there is an external force, so
$$ \textrm{d}U_1 + \textrm{d}U_2 \neq 0 $$
But we can also say that the operator doesn't touch the device. Then, $\delta W_{ext} = 0$, however the kinetic energy of the piston have changed during the process, so $\textrm{d}E_p \neq 0$, which gives
$$ \textrm{d}U_1 + \textrm{d}U_2 \neq 0$$
In both cases, at first sight the conservation of energy is violated, however once you've taken the piston into account, there's no problem anymore.
The correct answer to your question is simply that the changes in energy are not $P_1dV$ on one side and $P_2dV$ on the other side. The changes in energy are $-(P_1-P_2)dV$ on the high pressure side and $(P_1-P_2)dV$ on the low pressure side. There is no "extra energy" to account for.
Explanation: $dV$ is equal to $Adx$, where $A$ is the area of the piston and $dx$ is its displacement. The force on the high pressure side of the piston is $P_1A$, and the force on the low pressure side is $-P_2A$. The net force is the sum of the two, $(P_1-P_2)A$, and the energy transferred, or work done, is the product of this net force and the displacement $dx$. The work is done by the high pressure chamber on the low pressure chamber, so it is negative for the former and positive for the latter.
