I think you can try a variation with respect to $\omega_{ab}$. First of all $R^{ab} \neq D\omega^{ab}$, the correct statement is $$\delta R^{ab} = D\delta\omega^{ab}$$ Using this, we may do a variation of $E_4$ with respect to $\omega^{ab}$, to obtain

$$\delta E_4 = -2\int \epsilon_{abcd} D R^{ab}\wedge \delta\omega^{cd}$$ I have used integration by parts to shift the covariant exterior derivative on $R^{ab}$. However, already $DR^{ab} = 0$ by Bianchi identity (https://en.wikipedia.org/wiki/Curvature_form). So, this shows that of $E_4$ is independent of any variation in $\omega^{ab}$.

I think you can try a variation with respect to $\omega_{ab}$. First of all $R^{ab} \neq D\omega^{ab}$ because $\omega^{ab}$ is not a tensor, the correct statement actually is $$\delta R^{ab} = D\delta\omega^{ab}$$ Using this, we may do a variation of $E_4$ with respect to $\omega^{ab}$, to obtain

$$\delta E_4 = -2\int \epsilon_{abcd} D R^{ab}\wedge \delta\omega^{cd}$$ I have used integration by parts to shift the covariant exterior derivative on $R^{ab}$. However, already $DR^{ab} = 0$ by Bianchi identity (https://en.wikipedia.org/wiki/Curvature_form). So, this shows that of $E_4$ is independent of any variation in $\omega^{ab}$. Notice I have not used any assumption about torsion here. My proof is independent of that.