You missed just one detail in getting the right Thevenin resistance:
\$R_1\$, \$R_2\$, \$I_1\$, \$V_2\$, and \$R_5\$ can't alter anything as seen from A and B. What this subnet may alter is the current in \$V_1\$. But otherwise \$V_1\$ spans across that subnet and therefore bypasses it from the global perspective of terminals A and B.
Since you can use any method to get the Thevenin voltage, then I think Andy has set you up for that. You can either Nortonize the \$14\:\text{V}\$ source and the \$100\:\Omega\$ resistor, add that current source to \$I_2\$ and then Thevenize the resulting Norton. Or use some other approach.
The voltage at B will be more positive than the voltage at A, though. So be careful about specifying the resulting Thevenin voltage.