Thus, the air molecules contribute a small portion of their kinetic energy to the paddle, which is then expended as heat on the other side of the border, making the air molecules on the left colder, while air molecules on the right heat up. Doesn't this mean a decrease in entropy?
Yes it does.
However, we need to take the thermal noise of the resistor into account.
Hot resistors make noise
As discovered by John B. Johnson in 1928 and theoretically explained by Harry Nyquist, a resistor at temperature $T$ exhibits a non-zero open circuit voltage.
This voltage is stochastic and characterized by a (single sided) spectral density
$$S_V(f) = 4 k_b T R \frac{h f / k_b T}{\exp \left(h f / k_b T \right) - 1} \, . \tag{1}$$
At room temperature we find $k_b T / h = 6 \times 10^{12} \, \text{Hz}$, which is a ridiculously high frequency for electrical systems.
Therefore, for the loop of wire and resistor circuit in the device under consideration, we can roughly assume that
$$\exp(h f / k_b T) \approx 1 + h f /k_b T$$
so that
$$S_V(f) \approx 4 k_b T R \tag{2}$$
which we traditionally call the "Johnson noise" formula.
If we short circuit the resistor as in the diagram where its ends are connected by a simple wire, then the current noise spectral density is (just divide by $R^2$)
$$S_I(f) = 4 k_b T / R \, .\tag{3}$$
Another way to think about this is that the resistor generates random current which is Gaussian distributed with standard deviation $\sigma_I = \sqrt{4 k_b T B / R}$ where $B$ is the bandwidth of whatever circuit is connected to the resistor.
Johnson noise keeps the system in equilibrium
Anyway, the point is that the little resistor in the machine actually generates random currents in the wire!
These little currents cause the rod to twist back and forth for exactly the same reason that the twists in the rod induced by air molecules crashing into the paddles caused currents in the resistor (i.e. Faraday's law).
Therefore, the thermal noise of the resistor shakes the paddles and heats up the air.
So, while heat travels from the air on the left side to the resistor on the right, precisely the opposite process also occurs: heat travels from the resistor on the right to the air on the left.
The heat flow is always occurring in both directions.
By definition, in equilibrium the left-to-right flow has the same magnitude as the right-to-left flow and both sides just sit at equal temperature; no entropy flows from one side to the other.
Fluctuation-dissipation
Note that the resistor is both dissipative and noisy.
The resistance $R$ means that the resistor turns current/voltage into heat; the power dissipated by a resistor is
$$P = I^2 R = V^2 / R \, . \tag{4}$$
The noise is characterized by a spectral density given in Eq. (1).
Note the conspicuous appearance of the dissipation parameter $R$ in the spectral density.
This is no accident.
There is a profound link between dissipation and noise in all physical systems.
Using thermodynamics (or actually even quantum mechanics!) one can prove that any physical system which acts as a dissipator of energy must also be noisy.
The link between noisy fluctuations and dissipation is described by the fluctuation-dissipation theorem, which is one of the most interesting laws in all of physics.
The machine originally looked like it moved entropy from the left to the right because we assumed the resistor was dissipative without being noisy, but as explained via the fluctuation-dissipation theorem this is entirely impossible; all dissipative systems exhibit noisy fluctuations.
P.S. I really, really like this question.
Best Answer
This ratchet-like Maxwell's demon has the same problem as all of the other ones: the door/coil mechanism itself will heat up, and become useless.
Before thinking about this one, think about the simpler scenario where there's just a door, that opens if a fast particle hits it hard enough. Since particles have energy on the order of $kT$, the door must require around that much energy to open. But by the equipartition theorem, once the door itself is at temperature $T$, it will have $kT$ of thermal energy! So after a while the door will be wildly swinging open and shut on its own, and become totally useless.
This machine adds a second stage: now, your particle's $kT$ of kinetic energy goes into making a current in the coil, and the current in the coil opens the door. However, thermal noise applies to circuits, too; after a while, your coil will reach temperature $T$ and have $kT$ of Johnson noise, giving a randomly fluctuating current. As in the previous case, this will make your door randomly open and close, making the device fail.