The principle of relativity says that we can analyze a physical situation from any reference frame, as long as it moves with some constant speed relative to a known inertial frame. Thus, the ion drive does not find it more difficult to accelerate the ship when the ship is "going fast" because the ion drive cannot physically distinguish going fast from going slow.
However, if the ion drive is going fast in the reference frame of Earth, then when the ion drive burns, say 1 kg of fuel, it picks up less speed in the Earth frame than it does in the rocket frame due to the relativistic velocity addition law.
That velocity addition law is just the angle-addition law for the hyperbolic tangent. So, suppose the ship accelerates by shooting individual ions out the back. Each time it does this, it accelerates the same amount from its own comoving frame. Then from an Earth frame, the $\textrm{arctanh}$ of the rocket's speed increases by the same amount each time.
If, as a function of the proper time $\tau$ experienced on the rocket, the acceleration of the rocket is $a(\tau)$ in a comoving frame, there is a quantity called the rapidity of the rocket which increases the way velocity does in Newtonian mechanics.
The rapidity $\theta$ will be $\theta(\tau) = \int_0^\tau a(\tau) d\tau$, and the velocity is then $v(\tau) = \tanh\theta$. Specifically, if $a = g$, the velocity is
$$v(\tau) = \tanh(g\tau)$$
When one year of time has passed on the rocket, its velocity relative to Earth will be $\tanh(1.05) = 0.78$, or 78% the speed of light. The limit of the $\tanh$ function is one as $\tau \to \infty$, so the rocket never gets to light speed.
A more important limiting factor is the fuel. If the rocket carries all its fuel, then once it burns through it all, it can't go any more. Fusion isn't a way around this because by $E=mc^2$ there is a limited energy you can get from a given mass of fuel.
If a fraction $f$ of the rocket is fuel, when the fuel is all burned, the momentum of the rocket will be $\gamma m (1-f) \beta$, with $m$ the original mass. The energy of the rocket is $\gamma m (1-f)$. Similar relations hold for the fuel. The conservation of momentum and energy give
$$m = \gamma m (1-f) + E_{fuel}$$
$$0 = \gamma m \beta (1-f) + p_{fuel}$$
$E_{fuel}$ and $p_{fuel}$ are the energy and momentum of the fuel after burning. Solving for $\beta$ gives
$$\beta = \frac{-p_{fuel}}{m - E_{fuel}}$$
The minus sign shows that the fuel and rocket go opposite directions. To maximize $\beta$, we want to make $p_{fuel}$ as large as possible subject to a fixed $E_{fuel}$. This means that we want the speed of the fuel as high as possible, so assume the fuel is massless with $\beta_{fuel} = 1$ and $p_{fuel} = -E_{fuel}$. Plugging this into the previous equations and doing some algebra, I got
$$\beta = \frac{1 - (1-f)^2}{1 + (1-f)^2}$$
Even if half the rocket's original mass were fuel, it would only get to 3/5 the speed of light.
How can kinetic energy be proportional to the square of velocity, when
velocity is relative?
Without reading the rest of your question, I must first reply that one has nothing to do with the other.
Kinetic energy is frame dependent, just as velocity is.
Momentum is proportional to velocity and is frame dependent too, just as velocity is.
Now, looking at the body of your question:
Imagine you and your mate are in space drifting along together, at an
unknown speed.
Unknown speed relative to what? Unknown speed relative to Earth? Unknown speed relative to the solar system? Unknown speed relative to the CMB?
Assuming 100kg spaceships, will 5kJ of energy always produce 10m/s of
relative velocity?
Relative to what? Relative to the initial inertial frame of reference before the acceleration? Or relative to some frame of reference in some arbitrary relative motion?
(The point of all these questions is to prompt you to think more clearly about your question in the hope that you'll come to the answer yourself...)
Best Answer
I'm interpreting your question as you wanting to know the resultent velocity for an object of mass $m$ and initial velocity $v_0$ being accelerated in the direction of its velocity by a constant source of energy inputting at a constant rate $P$, the 'power'.
Since the power is the rate of change of energy, in this case kinetic energy, we have $$ P=\frac{dE} {dt} = \frac{1} {2} m \frac{dv^2} {dt} = m v \frac{dv} {dt} $$ thus $$ \int \frac{P} {m} dt = \int v dv $$ and $$ \frac{Pt} {m} = \frac{v^2} {2} +const $$
Applying the inital condition to find the constant, get that $$ v=\sqrt{v_0 + \frac{2Pt} {m}} $$
Does that help?