How do I modify a DC electric generator to get the same power at a lower speed?
Does increasing the number of turns by \$ \frac{2000}{700} \$ do the trick?


Almost. That should change the back-emf by a factor of K = 20/7, to compensate for the change in speed. The problem is that even if you manage to rewind the motor effectively, the electric machine's resistance and inductance will increase by a factor of K2 = 8.16 -- the I2R losses in the generator will increase by a factor of 8 at the same load current. And that's if you manage to rewind the motor effectively. If you can't reach a good fill factor on the rotor, the resistance will be even higher. You'll need proper equipment for this; I wouldn't try it by hand.

It's a good rule of thumb that the I2R losses in permanent magnet motors (whether synchronous or brush DC) are lowest at a given mechanical power level when the motors are running at higher speeds. Running them at lower speeds makes the torque requirements go up, and the I2R loss increases by this factor squared.

So if you can compensate for the I2R loss by making the winding ratio even higher, e.g. 21/7 or 22/7 (which causes even more I2R loss), and you don't overheat the generator, you'll meet your output power target.

This is why gears &amp; belts are often used with electric motors at low speeds, rather than using direct-drive.

The alternative approach is to make an electric machine with more poles: higher pole count = higher electrical frequency, which brings the electric machine's operating point closer to its area of highest efficiency. But that's more involved than just rewinding the motor.