Thevenin's theorem equivalent circuit
Your \$V_{th}\$ calculation is correct. But it's that funny that you don't see it cz of the notation \$V_{AB}\$ that you are using through out. You are trying to find the thevenin eq. circuit looking from the terminals A and B. And You have finally derived that:
$$V_{th} = -V_{AB}-1$$
But \$V_{AB}\$ is nothing but the open circuit voltage which is = \$V_{th}\$

So that makes the equation:
$$V_{th} = -V_{th}-1\implies V_{th} = -0.5V$$
However your  \$R_{th}\$ calculation is wrong. You just simply wrote down whatever resistor is left in the circuit. That's not the way to calculate it. Here is one method to find it:

In the first circuit, short all independent voltage sources, open all independent current sources and put an imaginary current source of 1A, across the terminals AB (across which you wanna calculate \$R_{th}\$).







simulate this circuit &ndash; Schematic created using CircuitLab

You can solve for  \$i_x= 1A\$ using KVL and Nodal analysis on V. Solve to get \$i_x= 1A\$

Finally, \$R_{th}\$ can be found out by calculating the voltage drop across our imaginary current source and applying Ohm's law:

$$R_{th} = V/i = V_{R2}/i = i_{x}R_2/1=1\Omega$$

You can now draw your thevenin ckt:







simulate this circuit