This method is almost correct (of course you have to remember that the final regular expression E is the union of E1 and E2), but there is something to adjust, if I remember properly: when you use the state elimination in more than one final state, when you remove one of them and keep the others you have to continue with state elimination method. For example, in the first automaton you eliminated q3 and keep q2, but before drawing initial->final states you need to calculate the arcs labels considering predecessors and successors as above. In this case they are both q0, R is its loop 1, Q=01, P=1, S=0 and so on.
At the end of this procedure, the arcs from q0 to q2 and viceversa are the same as you write correctly (I don't understand if the arc from q2 to q0 has + operator, remember that "0,1" means "0 OR 1" whichs is "0+1"), but the loop in q0 is quite more complex, 1+010*1
At the end of this procedure, the arcs from q0 to q2 and viceversa are the same as you write correctly (I don't understand if the arc from q2 to q0 has + operator, remember that "0,1" means "0 OR 1" whichs is "0+1"), but the loop in q0 is quite more complex, 1+010*1