Lets look at the use of extension tubes rather than a macro lens so that the effect becomes easier to visualize.
The effective f-stop of a lens is equal to the actual f-stop of the lens times (1 + magnification/pupil magnification). In lenses that are about 50mm or so, the pupil magnification is about 1. Longer lenses will have a smaller pupil magnification and shorter lenses will have a larger pupil magnification. For example, the Canon 180mm f/3.5L has a pupil magnification of 0.5 when focused at 1:1.
So, assuming a symmetric lens design with a pupil magnification of 1, we've got:
Fe = Fa * (1 + Magnification)
Now, if you've got that 50mm lens with 50mm of extension tubes, you are at a magnification of 1.0, and the effective f-stop (Fe) is twice the actual one. In other words, you've lost two stops of light in doing that. The lens system is indeed slower.
Look at it this way, the light is traveling twice the distance it did before to get to the media. The inverse square law then has it illuminating 4x the area (of which you only care about 1x of it) and that is again, 2 stops of light.
Note that this is still a 50mm lens in this example. Its just that you've traded a closer minimum focusing distance of the lens for the ability to focus at infinity.
I should point out that the example that I gave was with a nice, simple, symmetric lens that was used to do macro work.
When you've got internal focusing going on (rather than the old school 'move all the glass'), the simple lens equations are no longer simple, but many of the principles are still there, even when not working with a macro lens. The magnification of the subject changes, and the effective aperture changes along with it.