Colpitts oscillator: emitter resistor
Although you are correct that the emitter resistor will bypass some of the signal current the more important factor is that it affects the biasing of the transistor.

For an amplifier to become an oscillator one of the requirements is that there is feedback and that the overall loop gain is more than unity to account for the losses in various parts of the circuit.

Once oscillation is started the amplitude will keep increasing until something causes the gain to decrease so that it is exactly unity.  In this type of amplifier it can be caused by saturation of the amplifier where it is giving as much output as it can - this is usually related to the current flowing through the active device.

At that point the amplitude will be determined by the effective loss resistance of the resonant circuit and the output current of the amplifier.

By reducing the emitter resistor you are making the loss resistance worse by bypassing part of the signal current, but you are also increasing the quiescent current in the transistor and increasing its output current capability.  In your case the latter effect is larger than the former so the output amplitude is increased.

The transconductance of the transistor will also increase with current that will also change things and probably increase the gain.

You can calculate the DC conditions by ignoring the feedback and treating it as a simple amplifier.

With oscillators it may not be as simple as I have just described as there are also non-linear effects.  The AC signal at the emitter may get rectified by the base-emitter junction and affect the voltage at the base and so change the DC bias conditions - in many cases this may be the dominant factor, it can also cause the oscillations to be modulated at a low frequency where the oscillations will repeatedly stop and start, referred to as squegging. This behavior is often exploited in super regenerative receivers.