Yet, you can still consider U, F, and R (etc) as operations, correct? It is another way of thinking about it, perhaps, but still legitimate. It even makes more intuitive sense to consider it this way, as turns seem to be operations you do to a cube.ubervern, since you see U, F, R, etc. as "operations," you seem to be trying to relate them to an operation for a group. This seems to be why you don't see the natural relationship between Rubik's Cube and group theory.
More concretely, consider the cube a tensor with each element representative of a single cubie on the cube. All the cubies may be assigned a unique value that represents position and orientation. Then you may define U, F, R, etc as transformations which can be applied to your tensor through simple multiplication.
This will still work out to be associative, as tensor multiplication is. If you were to apply the algorithm U R F on state S, you would have: F*R*U*S. You can do (F*R)*U*S, and it will be the same as F*(R*U)*S.
Edit: Interestingly, as you only need to turn 5 faces to solve a cube, you will only need to use 5 operations. In (multi)linear algebra terms, you can show that the system of operations is dependent, and can thus determine that U is a combination of R, F, D, B, and L quite easily.
Edit again: Although, I just realized I'm not sure whether the transformations would be considered operations; could it be only the multiplication? Or the two combined? I think the two combined, as the transformations require both the transformation tensor and the multiplication definition to work.