Thevenine equivalent using loops
Everything is correct and your loops method work.  

The current \$I = \frac{3 V_S}{5 R}\$ therefore the \$R_{Th}\$ resistane is:

$$R_{Th} = \frac{V_S}{I} = \frac{V_S}{\frac{3 V_S}{5 R}} = \frac{V_S}{1} \cdot \frac{5R}{3V_S} = \frac{5R}{3} $$ 

Ans this is the resistance seen from the input voltage source point of view

 

And this is what you have found because you solve for the \$I\$ current (drawn from the voltage source).

But we can also find the resistance seen between \$A\$ and \$B\$ terminals: 



\$R_{th} = (0.5R+R)||R = 1.5R||R = 0.6R\$

But this resistance can be found as I have shown here:  

Why is the voltage of a capacitor equal to the voltage of a battery connected it?