I am afraid that the answer is:
I_norton = 10A
R_norton = 1 ohm
To find I_norton, we made a short (it is like placing a resistor of zero ohm) between terminals (a) and (b).
The current in this short is I_norton.
In this case, Vx=0 hence the dependent source 2Vx is also zero.
Therefore the source current (10A) will find its easiest path at the shorted branch only (no current will flow in the resistors 6 ohms and 2 ohms).
In other words:
I_norton = 10 A
To find R_norton, we do as you did
Let us write the equations.
Loop1:
I1*6 - 2*Vx + (I1-I2)*2 = 0 [equation 1]
Loop2:
(I2-I1)*2 - Vo = 0
-2*I1 + 2*I2 = Vo [equation 2]
But
Vx = Vo [equation 3]
Replacing Vx in [equation 1]:
I1*6 - 2*Vo + (I1-I2)*2 = 0
I1*6 + I1*2 - I2*2 = 2*Vo
8*I1 - 2*I2 = 2*Vo [equation 4]
Adding [equation 4] and [equation 2]:
6*I1 = 3*Vo
2*I1 = Vo [equation 5]
Replacing 2*I1 of [equation 5] in [equation 2]:
-Vo + 2*I2 = Vo
2*I2 = Vo + Vo
I2 = Vo [equation 6]
On this circuit:
R_norton = Vo / I2 = Vo / Vo = 1 ohm