Let $I$ be a ideal of $\Omega(M)$ ($M$ a manifold) that is locally generated by $p$ independent $1$-forms $\omega_1,...,\omega_p$. Set $\omega = \omega_1 \wedge ...\wedge \omega_p$ Then $I$ is a differential ideal iff $$d\omega = \alpha\wedge\omega $$for some $1$-form $\alpha$.
The $\implies$ direction is straightforward, since $d\omega_i = \sum \eta_j \wedge \omega_i$ for some $1$-forms $\eta$. The other direction has got me stuck. I thought about trying to prove that $d\omega_i \in I$, but I don't see how to use this hypothesis since it seems like I"m losing a lot of data.
Here's the idea for the proof. Locally, extend $\omega_1,\dots,\omega_p$ to a basis $\omega_1,\dots,\omega_p,\omega_{p+1},\dots,\omega_n$ for the $1$-forms on $M$. For simplicity, I'll take $p=2$ and $n=4$, but you can work out the general case with a bit more notation. Let $I=(\omega_1,\omega_2)$. Then, working mod $I$, we have \begin{align*} d\omega_1 &= f\omega_3\wedge\omega_4 \pmod I \\ d\omega_2 &= g\omega_3\wedge\omega_4 \pmod I. \end{align*} Now $d(\omega_1\wedge\omega_2) = f\omega_3\wedge\omega_4\wedge\omega_2 - g\omega_1\wedge\omega_3\wedge\omega_4$. This expression is of the form $\alpha\wedge\omega_1\wedge\omega_2$ for some $1$-form $\alpha$ if and only if $f=g=0$, as desired.