Edit: The answer to the questions above are that the two sets have the same cardinality. See the other answers for the answers to the specific questions you asked.
This answer shows how you can distinguish the sizes of the sets using a probability method.
Picking two random numbers independently from $(0,1)$, what is the probability that the entire interval between will be excluded from your set? This is only true if the two numbers "fail" at the same digits, so I think you get different probability measures this way, depending on your starting set. Essentially, you are excluding intervals from $[0,1]$ which add up to size $1$ in both your sets, but by excluding the squares of those intervals, you get a value less than $1$ and greater than $0$.
For example, in the case of your first set, if the two numbers you pick are in the interval $(0.5,0.6)$, then the entire interval between them is outside of your set. So the probability of picking a pair of numbers satisfying this condition is at least $0.01$, but cannot add up to $1$ because if $a\in (0.5,0.6)$ and $b\in(0.6,0.7)$ then there is always an element of your set between them.
I think this probability will measure these sets in the way you intend.
Edit: (From my comment below.) Okay, I don't like this answer, after all. For example, if you take the set $S_1$ to be the real numbers in (0,1) without a 5 or 6 in the decimal representation, you get a different value, from my method, than if you take the set $S_2$, the set with no 5 or 7 in the decimal representation. That strikes me as wrong - these two sets should not really be considered different in size, intuitively, but my computation comes up different for these two sets.