Find equivalent LTL formula, without Y (Yesterday) operator. How can I handle first state?
Initial states are somewhat of an anomaly. A similar problem is encountered in the $X$ operator when you try to define LTL over finite words. That is, it somehow would have made more sense to define PLTL over computations that are "infinite on both sides".

So, being an anomaly, it is probably easiest to treat it as such, and just explicitly deal with the initial state, requiring the first state not to have $a$. 

Thus, an equivalent formula would be $G(Xa\to b)\wedge \neg a$.