Showing that a transformation is measure preserving
To show that transformation is measure preserving you need to show that full preimage $S^{-1}(A)$ of any set A in the Borel $\sigma$-field on [0,1) is again in the same $\sigma$-field, i.e. it is measurable, and that $Pr\{S^{-1}(A)\} = Pr\{A\}.$

As you have correctly pointed out, it will suffice to show it for any base of usual topology, namely, all open intervals.

It is evident that preimage of $(a,b)$ is $(\frac{a}{2},\frac{b}{2})\cup (\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2})$. It is, evidently, in Borel $\sigma$-field, as it is a union of two open intervals, and $Pr\{(\frac{a}{2},\frac{b}{2})\cup (\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2})\} = Pr\{(a,b)\} = b-a$

We proved that preimage of any open interval is measurable and has the same measure as the interval. As open intervals form a base of the usual topology we can extend this result to any measurable set and it completes the proof.