General Distributive Law and Axiom of Choice

I don’t know a reference, but I can give you a fairly short proof. (I want it available as a reference for myself, even if you really want only a reference.)

The harder direction is that the general distributive law in question implies $\mathsf{AC}$. I’ll prove $\mathsf{AC}$ in the following form: for any sets $X$ and $Y$, each surjection $f:X\to Y$ has a right inverse.

Proof: Let $X$ and $Y$ be sets, and let $f:X\to Y$ be surjective. For $\langle y,x\rangle\in Y\times X$ let $$A(y,x)=\begin{cases}\{0\},&\text{if } y=f(x)\\0,&\text{otherwise}\;.\end{cases}$$ For each $y\in Y$ there is an $x\in X$ such that $y=f(x)$, so $\bigcup_{x\in X}A(y,x)=\{0\}$, and therefore $$\bigcup_{g\in{^YX}}\bigcap_{y\in Y}A\big(y,g(y)\big)=\bigcap_{y\in Y}\bigcup_{x\in X}A(y,x)=\{0\}\;.$$ Thus, there is a $g\in{^YX}$ such that $$\bigcap_{y\in Y}A\big(y,g(y)\big)=\{0\}\;,$$ i.e., such that $y=f\big(g(y)\big)$ for each $y\in Y$. Clearly $f\circ g=\operatorname{id}_Y$, as desired. $\dashv$.

The other direction is routine. Assume $\mathsf{AC}$. Given sets $A(i,j)$ for $\langle i,j\rangle\in I\times J$, suppose that $x\in\bigcap_{i\in I}\bigcup_{j\in J}A(i,j)$; then $J_x(i)=\{j\in J:x\in A(i,j)\}\ne\varnothing$ for each $i\in I$. Let $\xi:I\to J$ be a choice function for $\{J_x(i):i\in I\}$; then

$$x\in\bigcap_{i\in I}A\big(i,\xi(i)\big)\subseteq\bigcup_{\varphi\in{^IJ}}\bigcap_{i\in I}A\big(i,\varphi(i)\big)\;.$$ On the other hand, if $x\in\bigcup_{\varphi\in{^IJ}}\bigcap_{i\in I}A\big(i,\varphi(i)\big)$, then there is a $\varphi\in{^IJ}$ such that

$$x\in\bigcap_{i\in I}A\big(i,\varphi(i)\big)\subseteq\bigcap_{i\in I}\bigcup_{j\in J}A(i,j)\;.$$