I think this works. We prove a slightly stronger statement by induction on $\dim X$: Let $X$ be a quasi-projective variety over $\mathbb{C}$. Let $x_1$, $x_2$, ..., $x_n$ be units in $H^0(X, \mathcal{O})$. By "$d \log x_i$" I mean the class in $H^1(X)$ pulled back from the generator of $H^1(\mathbb{G}_m)$ by the map $x_i : X \to \mathbb{G}_m$. (I say this to address the possibility that $X$ isn't smooth, since I'm not sure whether Kahler differentials give cohomology classes in general.) Then, as before, I claim that:
If a polynomial $\eta$ in the $d \log x_i$ is exact, then it is $0$.
The base case, $\dim X=0$, is trivial.
Reduction We may (and do) assume that $X$ is smooth.
Proof Let $\tilde{X} \to X$ be a resolution of singularities. Since the class of $\eta$ is $0$ in $H^{\ast}(X)$, it pulls back to $0$ in $H^{\ast}(\tilde{X})$. So $\eta$ pulls back to $0$ on $\tilde{X}$, and is thus $0$. $\square$
Now, choose a simple normal crossing compactification $\bar{X}$ of $X$. Let $D_1$, $D_2$, ..., $D_m$ be the components of $D$. Our next goal is to prove
Claim For any component $D_1$ of $D$, the form $\eta$ has no pole on $D_1$.
Reduction We may assume that $x_2$, $x_3$, ...., $x_n$ don't have poles or zeroes along $D_1$, and $x_1$ vanishes on $D_1$.
Proof Let $d_i$ be the order to which $x_i$ vanishes on $D_1$. Let $d = GCD(d_1, d_2, \dots, d_m)$. Then by applying a monomial transformation to the $x$'s, we may assume that $x_1$ vanishes to order $d$ on $D_1$. $\square$
Let $Y = D_1 \setminus \bigcup_{i \geq 2} D_1 \cap D_i$ and let $Z$ be the union of $X$ and $Y$ inside $\bar{X}$. So $Y$ is a smooth hypersurface in $Z$. We claim that the $x_i$ extend to homolomorphic function on $Z$, and that $x_2$, ..., $x_n$ extend to nonvanishing functions. Proof: Suppose $x_i$ does not extend to $Z$. Then $x_i$ has a pole along some hypersurface in $Z$. By the previous Reduction, that hypersurface is not $Y$, so it must meet $X$. But $x_i$ is well defined on $X$. For $i \geq 2$, this argument also shows that $x_i^{-1}$ extends.
Proof of Claim Let $\omega = \mathrm{Res}_{Y} \eta$. The form $\omega$ is a $d \log$ form on $Y$, in the ring generated by $d \log x_2$, $d \log x_3$, ..., $d \log x_n$. The residue map is a well defined map $H^k(X) \to H^{k-1}(Y)$. More specifically, let $U$ be a tubular neighborhood of $Y$, so $\partial U \subset X$ is a circle bundle over $Y$. Then $\mathrm{Res}$ is the composition of restriction $H^k(X) \to H^k(\partial U)$ and the Gysin map $H^k(\partial U) \to H^{k-1}(Y)$.
So $\omega$ is exact on $Y$. By induction, $\omega=0$. We have now established that the residue of $\eta$ along $Y$ is $0$. But $\eta$ clearly has at most a first order pole on $Y$, so it has no pole at all. $\square$.
Proof of theorem We have shown that $\eta$, as a form on $\bar{X}$, does not blow up on any of $D_i$. Since $\bar{X}$ is smooth, this shows that $\eta$ extends to $\bar{X}$. But, by Hodge theory, on a smooth projective variety, any exact holomorphic form is $0$.
Alternate proof As above, reduce to $X$ smooth; let $\overline{X}$ be a normal crossing compactification; let $D = \overline{X} \setminus X$. Let $\eta$ be a $k$-form. The condition that $\eta$ is generated by $d \log$ forms implies that $\eta$ is an element of $H^0(\overline{X}, \Omega^k(\log D))$. The condition that $\eta$ is exact says that the image of $\eta$ in $H^k(X, \mathbb{C})$ is $0$.
Conveniently, there is a spectral sequence $H^q(\overline{X}, \Omega^p(\log D)) \Rightarrow H^{p+q}(X, \mathbb{C})$ which degenerates at $E_1$. (I'm looking at Voisin's book, volume 1, Theorem 8.35, she cites Deligne Theorie de Hodge II.) So $H^0(\overline{X}, \Omega^k(\log D))$ injects into $H^k(X, \mathbb{C})$, which was the desired claim.
Thomas Lam and I have finally started writing the paper where we needed this lemma; this is probably the proof we will use.
