The answer to my own question. Many thanks to Jesse Peterson for pointing out the confusing place.

Let $\mathcal A$, $\mathcal B$ be sigma-algebras of subsets of $X$ and $Y$ respectively. Define the following two algebras of subsets on $X\times Y$

- $\mathcal A \times \mathcal B$ be an algebra of all step-sets: finite unions of disjoint subsets of the form $A\times B$, $A\in \mathcal A$, $B\in \mathcal B$ ("measurable rectangles").
- $\mathcal A \otimes \mathcal B$ be a $\sigma$-algebra generated by the algebra of step-sets.

When we restrict a normalized state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ to indicators of step-sets, we obtain a probability finitely-additive measure on the algebra $\mathcal A \times \mathcal B$ (due to positivity and the normalizing condition of a state). Its restrictions on $\mathcal A$, $\mathcal B$ appear to be countably additive absolutely continuous measures due to binormality: countable additivity corresponds normality, and absolute continuity follows from the Riesz representation of $(L^\infty(\mu_X))^*$ as the space of bounded absolutely continuous finitely-additive measures.

Then we should extend our finitely-additive measure from $\mathcal A \times \mathcal B$ to $\mathcal A \otimes \mathcal B$. It is not hard to prove, that countable additivity of marginals implies countable additivity of our measure on $\mathcal A \times \mathcal B$ (see "Lemma" below). By Hahn-Kolmogorov theorem, it has the unique countably-additive extension on $\mathcal A \otimes \mathcal B$.

It is straightforward to check that any measure of the specified type defines a binormal state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ via integration.

**Lemma**.
*Any finitely additive measure $\gamma$ on $\mathcal A \times \mathcal B$ with countably additive marginals is countably additive on $\mathcal A \times \mathcal B$.*

*Proof.* Since $\gamma$ is finitely additive, it follows that for a countable family $\{C_i\}$ of disjoint elements of $\mathcal A\times \mathcal B$ with $\bigcup_{i=1}^\infty C_i\in \mathcal A\times \mathcal B$
$$
\gamma\left(\bigcup_{i=1}^\infty C_i\right)\geq \sum_{i=1}^\infty\gamma(C_i)
$$
Note that for a countable family $\{A_k\}$ of disjoint elements of $\mathcal A$ and any $B\in\mathcal B$ we have
$$
\gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)-\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)
$$
Then we can use countable additivity of the marginal
$$
\gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)+\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)=\sum_{k=1}^\infty\gamma(A_k\times Y)
$$
Since it is forbidden for $\gamma$ to have a strict inequality of the form
$$
\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)< \sum_{i=1}^\infty\gamma(A_k\times (Y\setminus B))
$$
we conclude that
$$
\gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)= \sum_{k=1}^\infty\gamma(A_k\times B)
$$
and hence all measures of the form $\mu_B(A):=\gamma(A\times B)$, $B\in \mathcal B$ on $\mathcal A$ are countably additive. By analogous argument all measures $\nu_A(B):=\gamma(A\times B)$, $A\in \mathcal A$ on $\mathcal B$ are also countably additive.

Let $\{A_n\}$ be a countable family of disjoint elements of $\mathcal A$, $\{B_k\}$ be a countable family of disjoint elements of $\mathcal B$, $A=\bigcup_{n=1}^\infty A_n$, $B=\bigcup_{k=1}^\infty B_k$, $\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty (A_n\times B_k)\in \mathcal A \times \mathcal B$, then
$$
\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty A\times B_k\right)=\sum_{n=1}^\infty\gamma(A\times B_k)=\sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k)
$$
Since
$$
\sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)+\gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right)
$$
and
$$
\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)\geq \sum_{k=1}^\infty\sum_{n\neq k}\gamma(A_n\times B_k)
$$
we conclude that
$$
\gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right)=\sum_{n=1}^\infty\gamma(A_n\times B_n)
$$

Let $C=\bigcup_{n=1}^\infty D_n$, $C=\bigcup_{j=1}^N C_j$, $D_n=\bigcup_{i=1}^{M_n} D_{n,i}$, where the families $\{C_j\}$ and $\{D_{n,i}\}$ are families of disjoint measurable rectangles. It follows that $D_{n,i,j}=D_{n,i}\cap C_j$ is also a family of disjoint measurable rectangles. Since $C_j=\bigcup_{n=1}^\infty \bigcup_{i=1}^{M_n} D_{n,i,j}$, we can use the obtained result to show that
$$
\gamma(C_j)=\sum_{n=1}^\infty \sum_{i=1}^{M_n} \gamma(D_{n,i,j})
$$
Since $D_{n,j}=\bigcup_{j=1}^{N} D_{n,i,j}$, it is obvious that $\gamma(D_{n,j})=\sum_{j=1}^N \gamma(D_{n,i,j})$. It is also true that $\gamma(C)=\sum_{j=1}^N \gamma(C_j)$, and it follows that
$$
\gamma(C)=\sum_{i=1}^\infty \gamma(D_n)
$$
which completes the proof of the lemma.