The energy involved in precession has its origin in the kinetic energy in the spinning gyroscope. So far so good.
No.
You spin the gyroscope (upright) so that it has an angular velocity $\omega$. The kinetic energy associated with this motion is:
$$ \frac{1}{2}I_{\mathrm{axle}}\omega^2,$$
where $I_{\mathrm{axle}}$ is the moment of intertia of the gyro about its axle.
The gyroscope then falls a bit under gravity and then it starts precessing. The angular velocity of the gyroscope about its axle is still $\omega$ -- assuming a point contact with the ground and no air resistance, there was no torque in that direction that could have reduced it.
The gyroscope then starts precessing about the $z$ axis (gravity) with angular velocity $\Omega$. This introduces a kinetic energy $$ \frac{1}{2}I_z \Omega^2,$$
where $I_z$ is the momentum of intertia of the gyro assembly about the $z$ axis, not about its axle.
This kinetic energy, not originally present in the system, comes form the gravitational potential energy $mgL(1-\cos\theta)$ lost when the gyro drops by angle $\theta$ before precessing.
For the angular momentum of the precession (not of the gyroscope wheel
itself) to be preserved, there must be an equal and opposite pointing
angular momentum somewhere in the system.
Angular momentum (about an axis and/or a point) is only conserved in a closed system, i.e. a system which no external torques act upon.
For the $\omega$-associated angular momentum, that's still conserved as said before, there are no torques tangential to the gyro that can slow down its rotation.
For the total angular momentum, about (say) the point of contact, gravity provides the external torque.
Hence you expect a change in angular momentum:
$$ \Delta \mathbf{L} = \Gamma \Delta t = (\mathbf{r}\times m\mathbf{g})\Delta t,$$
which is exactly what causes precession.