I think you are wrong; for the given structure, we first get the following relation
step 0: {(S0,S0);(S0,S3);
(S1,S1);(S1,S2);(S1,S4);(S1,S5);
(S2,S1);(S2,S2);(S2,S4);(S2,S5);
(S3,S0);(S3,S3);
(S4,S1);(S4,S2);(S4,S4);(S4,S5);
(S5,S1);(S5,S2);(S5,S4);(S5,S5)}
These are the set of pairs of states with the same labels. Now, we have to check the simulation diagrams:
S0 > {S1;S2}

S3 > {S2;S4}
S1 > {S2}

S2 > {S2;S3}
S1 > {S2}

S4 > {S2;S5}
S1 > {S2}

S5 > {S0;S2}
S2 > {S2;S3}

S4 > {S2;S5}
S2 > {S2;S3}

S5 > {S0;S2}
S4 > {S2;S5}

S5 > {S0;S2}
To check for BISIM2a, for every transition in the upper row, we have to find one in the lower row, and for BISIM2b, for every transition in the lower row, we have to find a corresponding one in the upper row. The following pairs of states remain:
step 1: {(S0,S0);(S0,S3);
(S1,S1);(S1,S4);
(S2,S2);(S2,S5);
(S3,S0);(S3,S3);
(S4,S1);(S4,S4);
(S5,S2);(S5,S5)}
This is the relation B* that is the greatest relation satisfying BISIM1, BISIM2a and BISIM2b. It is clearly a bisimulation relation since it contains the identity relation. Hence, we can compress the structure to only three states.