Doesn't relativistic mass increase mean increased energy released when matter is turned to energy?

If you define $gamma m$ as the relativistic mass, the total energy of a lump of fuel with rest mass $m$ is $E=gamma mc^2$. After turning the mass into energy we have $E=pc$ where $p$ is the magnitude of the 3-momentum that can be transferred to the ship. But no matter how much the ship's momentum is increased, it will never reach the speed of light, since most of the momentum increase will only contribute to the ship's $gamma$ factor.

It doesn't matter what method is used to accelerate the ship, it won't reach the speed of light even if you send in extra fuel or momentum-carrying particles from outside.

Side note: The full formula for the energy of a moving particle is $E=sqrt{m^2c^4+p^2c^2}$. From this you can derive $E=mc^2$ for the special case of a stationary particle, and $E=pc$ for the special case of a massless particle. It's easy to verify that $gamma mc^2$ is the same as $sqrt{m^2c^4+p^2c^2}$, but the latter is much more helpful. It's better to always think of "mass" as the rest mass, since it is an invariant of motion.

If you're defining relativistic mass as $gamma m$ (an old and misleading consideration), then that is the "mass" determined from the observer frame outside the ship and moving relative to the mass, but the mass isn't $gamma m$ in the rest frame of the ship. The annihilation takes place in the ship frame where the mass of each particle is still only 511 keV/c$^2$.

What the outside observer will observe, moving with speed $beta$ compared to the ship and its propulsion unit, is that the photons are Doppler shifted to different wavelengths and different momenta (both travelling at $c$ in opposite directions) so that the momentum of each electron/positron pair is conserved (now with $p=2gamma mbeta c$ compared to zero in the ship.)