Demystifier said:
I think you are doing a category mistake here. You cannot add together forces that belong to different categories. The force of electric field belongs to the category of
fundamental microscopic forces, together with gravitational, weak and strong force. The force of rocket belongs to the category of forces caused by various macroscopic objects, such as rocket, Earth, hammer, etc. The force of rocket on the charge is of the electromagnetic origin, so you are doing a kind of double counting. (Which reminds me of the "proof" that horse has 8 legs; 2 left legs, 2 right legs, 2 forward legs and 2 backward legs.

)
This is not actually clarifying anything. Let me try to explain again.
Let's pick an inertial coordinate system. Suppose that at time ##t_0##, you have a situation where there is a region of space that can be described in this coordinate system by:
- There is a charged particle with mass ##M## and charge ##Q##.
- The particle have 3-velocity ##\vec{v}##.
- There is an electromagnetic field ##F^{\mu \nu}##.
I think everyone would agree that in accounting for energy conservation, you can't just use ##E = \gamma M c^2##. The total energy must also take into account electromagnetic energy.
So if you replace the charged particle by an uncharged particle, and keep the 3-velocity ##\vec{v}## the same and keep the mass ##M## the same, the total energy will not be the same, because the electromagnetic energy will be different. Does anybody disagree with this?
If you agree, then it follows that you can't just use ##E = \gamma M c^2## to calculate how much energy was needed to accelerate the mass ##M## to achieve velocity ##\vec{v}##. The charge ##Q## and the electromagnetic field ##F^{\mu \nu}## must be taken into account, as well.
If it turns out that the energy required to accelerate a particle from rest to velocity ##\vec{v}## is independent of whether it's charged, that's a pretty remarkable thing. I'm pretty sure it's not true, in general.
Now, instead of looking at the general case, let's look at the special case where ##F^{\mu \nu}## is the electromagnetic field due to the charge ##Q##. In that case, maybe you can say that some of the electromagnetic energy is already accounted for in the mass ##M##. That is, the mass ##M## already takes into account the electromagnetic self-energy. Does it then follow that the amount of energy required to accelerate a charged particle is the same as for an uncharged particle of the same mass? Maybe that's true, but it certainly isn't true by definition. It requires an argument.