Motivation
Consider a finite set $[q]=\{1,\dots,q\}$, random variables $X_1,\dots,X_k\in[q]$, and their product $X=X_1\otimes\cdots\otimes X_k\in[q]^k$, i.e. the components of $X$ are independent. Let $r:[q]^k\rightarrow[0,\infty)$ be such that $\mathbb E[r(X)]=1$, i.e. $r$ is a Radon-Nikodym derivative, and let $X^*\in[q]^k$ be given by the corresponding distribution, i.e. $\mathbb P(X^*=x)=\mathbb E[r(X)\unicode{120793}\{X=x\}]=r(x)\mathbb P(X=x)$.
Claim: We have $D(X^*\|X)-D(X^*_1\otimes\cdots\otimes X^*_k\|X)=D(X^*\|X^*_1\otimes\cdots\otimes X^*_k)$, where $D(A\|B)=\sum_a\mathbb P(A=a)\ln(\mathbb P(A=a)/\mathbb P(B=a))$ is the relative entropy.
Proof: We have $D(X^*\|X)-D(X^*_1\otimes\cdots\otimes X^*_k\|X)=H(X^*\|X)-H(X^*)-\sum_{h=1}^k[H(X^*_h\|X_h)-H(X^*_h)]=\sum_{h=1}^kH(X^*_h)-H(X^*)=D(X^*\|X^*_1\otimes\cdots\otimes X^*_k)$, where $H(A\|B)=-\sum_a\mathbb P(A=a)\ln(\mathbb P(B=a))$ is the cross entropy and $H(A)=H(A\|A)$ is the entropy.
The quantity $D(X^*\|X^*_1\otimes\cdots\otimes X^*_k)$ is known as total correlation.
General Setup
For any random objects $X_1\in\mathcal X_1,\dots,X_k\in\mathcal X_k$ let $X=X_1\otimes\cdots\otimes X_k\in\mathcal X=\prod_{h=1}^k\mathcal X_h$ be the product, let $r:\mathcal X\rightarrow[0,\infty)$ be such that $\mathbb E[r(X)]=1$, and let $X^*$ be given by the Radon-Nikodym derivative with respect to $X$.
Question: Is it true that $D(X^*\|X)-D(X^*_1\otimes\cdots\otimes X^*_k\|X)=D(X^*\|X^*_1\otimes\cdots\otimes X^*_k)$, where $D(A\|B)=\mathbb E[\ln(r(A))]=\mathbb E[r(B)\ln(r(B))]$ is the general relative entropy if $A$ is given by the Radon-Nikodym derivative $r$ with respect to $B$.
If this does not hold in general, then the follow-up question is in what generality it does hold? I believe that it holds in general and that the proof is a one-liner as above. Even if that's not the case, this has to hold in greater generality. Say, for compact subsets of $\mathbb R^d$ you could approximate the measures with finitely supported ones.
It's exactly the same thing, but you have to invoke a little bit of calculus of RN derivatives to justify. Let $Q$ denote the law of $X^*$, $Q_i$ the law of $X_i^*$, and $P_i$ the law of $X_i$. I'll denote the product measures as $(\otimes Q)$ and $P = (\otimes P).$ Note that $ r(x) = \frac{\mathrm{d} Q}{\mathrm{d}P}(x).$ Let $x_{-i}$ denote $(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_n)$. We have, by definition, that \begin{align*} Q_i(E) &= \int_E \int r(x) (\otimes_{j \neq i} P_j)(\mathrm{\mathrm{d}}x_{-i})P_i(\mathrm d x_i) \implies\\ r_i(x_i) &:= \frac{\mathrm{d}Q_i}{\mathrm{d}P_i}(x_i) = \int r(x)(\otimes_{j \neq i} P_j)(\mathrm{\mathrm{d}}x_{-i}). \end{align*}
But then note that $$ \frac{\mathrm{d}(\otimes Q_j)}{\mathrm{d}(\otimes P_j)}(x) = \prod r_i(x_i) =: \tilde{r}(x).$$
Now, $$ D(Q\|P) = \int r(x) \log r(x) P(\mathrm{d}x),$$ and $$ D((\otimes Q_j)\|(\otimes P_j)) = \int \tilde{r}(x) \log\tilde{r}(x) P(\mathrm{d}x) = \sum_i \int \tilde{r}(x) \log r_i(x_i) P(\mathrm{d}x)\\ \overset{a}= \sum_i \int r_i(x) \log r_i(x) P_i(\mathrm{d}x_i) \overset{b}= \sum_i \int r(x) \log r_i(x) P(\mathrm{d}x)\\ = \int r(x) \log \tilde{r}(x) P(\mathrm{d}x),$$ where $a$ is by marginalising, and $b$ is by noting that for each $i$, the $i$th marginal laws of $Q$ and $(\otimes Q_j)$ both agree, and we are repeatedly using that $P = (\otimes P_j)$.
But now we can write that $$ D(Q\|P) - D((\otimes Q_j)\|P) = \int r(x) \log \frac{r(x)}{\tilde{r}(x)} P(\mathrm{d}x).$$ But note that if $ \mu \ll \nu \ll \mu,$ then $\frac{\mathrm{d}\mu}{\mathrm{d}\nu} \cdot \frac{\mathrm{d}\nu}{\mathrm{d}\mu} = 1.$ Using this, we can write $$ \int r(x) \log \frac{r(x)}{\tilde{r}(x)} (\otimes P_j)(\mathrm{d}x) = \int \frac{r(x)}{\tilde{r}(x)}\log \frac{r(x)}{\tilde{r}(x)} (\otimes Q_j)(\mathrm{d}x).$$ This is seen to equal $D(Q\|\otimes Q_j)$ upon observing that if $\mu \ll \lambda \ll \nu,$ then $ \frac{\mathrm{d} \mu}{\mathrm{d}\nu} = \frac{\mathrm{d}\mu}{\mathrm{d}\lambda} \cdot \frac{\mathrm{d}\lambda}{\mathrm{d}\nu}.$ Take $\mu = Q, \lambda = (\otimes P_j), \nu = (\otimes Q_j),$ and use our previous observation that $1/\tilde{r} = \mathrm{d}(\otimes P_j)/\mathrm{d}(\otimes Q_j)$
For this to formally work, you need $a,b$ to hold, and all the R-N derivatives to make sense. It should be enough to work in a Polish space (which is plenty general enough for most things).