Actually, as you have written it, the ring $R$ is generated by $\omega_0$ and the real and imaginary parts of $\Omega_0$, since $R$ consists of real-valued $1$-forms. You should probably consider, instead the ring $R^\mathbb{C} = \mathbb{C}\otimes R$ of complex-valued exterior forms on $\mathbb{R}^{2n}=\mathbb{C}^n$. Then the proof goes in two stages:

First, consider the forms that are invariant under the diagonal maximal torus in $\mathrm{SU}(n)$, which acts as $g\cdot dz^i = e^{\lambda_i} dz^i$ and $g\cdot d\bar z^i = e^{-\lambda_i} d\bar z^i$ where $\lambda_i$ are purely imaginary and satisfy $\lambda_1+\cdots+\lambda_n=0$. It is easy to see that the only complex-valued forms that are invariant under this torus action are the ones generated by $\omega_i = dz^i\wedge d\bar z^i$ and $\Omega_0$ and $\overline{\Omega_0}$. In particular, this shows that $R$ lies in the ring generated by the $(1,1)$-forms and $\Omega_0$ and $\overline{\Omega_0}$. Note that $\Omega_0$ (and $\overline{\Omega_0}$, too) wedged with any $(1,1)$-form must vanish, so it's clear that, outside of $\Omega_0$ and $\overline{\Omega_0}$, the ring $R$ must lie in the ring that is the sum of the $(p,p)$-forms.

Now, each $(p,p)$-form $\Psi$ can be written uniquely in the form
$$
\Psi = \Psi_p + \omega_0\wedge\Psi_{p-1} + {\omega_0}^2\wedge\Psi_{p-2} + \cdots + {\omega_0}^p\wedge\Psi_0
$$
where each $\Psi_j$ is an $\omega_0$-*primitive* $(j,j)$-form, and these primitive subspaces are $\mathrm{SU}(n)$-irreducible. (This follows from a root space representation.) In particular, if $\Psi$ is invariant under $\mathrm{SU}(n)$, then $\Psi = {\omega_0}^p\wedge \Psi_0$, so $\Psi$ is a complex multiple of ${\omega_0}^p$.