Thevenin equivalent of the following circuit
I already understood my misinterpretation of the question before. vo is the output voltage (measured between A and B) and not the voltage through the resistor. Therefore, the dependent source depends on the voltage between A and B and not on the voltage through the resistor.

That should not cause any trouble because what happens in the open-circuit test is that \$ v_O\$ will be equal to \$ V_{Th}\$ (it means, it's equal to the open-circuit voltage, in this case). Therefore we simply apply KVL and ohm's law:

$$V_{Th} = R_2 * I$$
$$R_1*I + a*V_{Th} + R_2*I - v_1 =0$$

After doing some manipulation

$$Vth=\frac{R_2}{R_1 + (1+a) R_2}*v_1$$

Since the equivalent resistance was correct, I'll reproduce my previous resolution:

Now for the equivalente resistance, I'm going to attach to AB a voltage source equal to \$v_o\$. A current \$i_o\$ will flow to the circuit and my ratio \$v_o/i_o\$ will be the equivalent resistance. Don't forget of course to short-circuit the independent voltage source. Applying KVL and KCL:

$$- v_0 - a v_0 + R_1 * i_1=0$$
$$v_0=R_2*i_2$$
$$i_0=i_1+i_2$$

And therefore

$$v_0=\frac{R_1*i_1}{1+a}$$
$$i_0=i_1+\frac{v_0}{R_2}$$

And after some manipulation:

$$i_1=\frac{v_0*(1+a)}{R_1}$$
$$i_0=\frac{v_0*(1+a)}{R_1}+ \frac{v_0}{R_2}$$
$$i_0=\frac{R_2(1+a)+R_1}{R_1*R_2}*v_0$$
$$R_{eq}=\frac{v_0}{i_0}=\frac{R_1*R_2}{R_2(1+a)+R_1}$$

Ok, now you have your Thèvenin circuit, computing the output voltage must be easy. Sorry for the confusion, I wrongly assumed that the voltage vO was the voltage across the resistor (and therefore the dependent voltage source depended on the voltage across the resistor) which is not the case. Good luck!