Do any non-inertial "forces" [terms in the metric] (like Coriolis in Newtonian mechanics) appear to a rotating observer (reference frame) in special relativity? Is the resulting spacetime after performing a change of coordinates in Cartesian Minkowski metric still FLAT, i.e., is it still special (and not general) relativity?
The simple answer is to point out that the Riemann curvature tensor and its various contractions are invariants i.e. they do not depend on the coordinates that you are using. For a non-rotating observer in flat spacetime the Riemann tensor vanishes everywhere, and therefore for the rotating observer the Riemann tensor must also vanish everywhere. Spacetime is flat for both observers.
Calculating the Minkowski metric in rotating polar coordinates has proved a bit much for me on a Friday evening, but can I suggest a similar comparison? The metric of a uniformly accelerating observer is the Rindler metric, and this metric has horizons like a Schwarzschild black hole (and in fact a Schwarzschild white hole). Despite this the spacetime is still flat and the Riemann tensor vanishes everywhere, as you can show with a simple coordinate transformation.
Special relativity applies to inertial reference frames, not rotating ones. For example, lets imagine your eye is the reference frame, you look at a star in the sky, then in a fraction of a second you rotate your eye to look away from the star. From your eye reference frame, the star moved faster than the speed of light.