When you add the 1 volt test voltage at terminals a-b, you no longer have 4 meshes; you now have 5.
Your top schematic shows 4 meshes, and this allows you to calculate Vth, because the 2Ω resistor has no effect when the output is open circuited. Vth is just the voltage across the 8Ω resistor.
But as soon as you add the 1 volt test voltage as in the bottom schematic, you have 5 meshes. You need the current in that 5th mesh to calculate Rth.
You could solve it with only 4 meshes by assuming a short across a-b; this then replaces the 8Ω resistor with the parallel combination of 2Ω and 8Ω. Solving your 4 equations would give you the current through that parallel combination. Then use the current divider rule to calculate the current through the 2Ω resistor; that would be the short circuit current through a-b.
If you solve this using the nodal method, you only have 3 equations to deal with. That's what I meant when I said "You usually can use any network analysis method. One method may be easier for a given circuit than another."