No that does not really make sense to me, maybe I got something wrong. First of all, it is not the case that [a WU b] is equivalent to (F b) | (G a), why do you think so?
I am also not sure whether you are drawing one or two structures since there are two states with label s0.
The path with the self-loop on the second state s0 satisfies G(a&b) and therefore also [a WU b], but it does not satisfy F!a. However, it does also not satisfy [b SB !a] since b does not hold before a, but at the same time.
The other path starting in the left state s0 satisfies [a WU b] since b immediately holds, and it also satisfies F!a since in s1, !a holds. Hence, it satisfies S1. It does not satisfy S2 so it is opposite to what you wrote.