Assume that
$E \in M_1 \times M_2$ such that
$\mu_1\times \mu_2(E)=0$. We have to show that $\nu_1\times \nu_2(E)=0$.
By Fubini theorem we have
$$0=\mu_1\times \mu_2(E)=\int_{X_1}\mu_2(E \cap (\{x\} \times X_2))d\mu_1(x).$$
This means that
$$
\mu_{1}(\{ x : \mu_2(E \cap (\{x\} \times X_2))>0 \})=0,
$$
equivalently
$$
\mu_{1}( X_1 \setminus \{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \})=0.
$$
Since $\nu_2 \ll \mu_2$ we have
$$
\{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \} \subseteq \{ x : \nu_2(E \cap (\{x\} \times X_2))=0 \}.
$$
Since $\nu_1 \ll \mu_1$ and
$$
\mu_{1}( X_1 \setminus \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=0,
$$
we have
$$
\nu_{1}( X_1 \setminus \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=\nu_{1}(\{ x : \nu_2(E\cap (\{x\} \times X_2))>0\})=0.
$$
Finally, we get
$$\nu_1\times \nu_2(E)=\int_{X_1}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=
$$
$$
\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))>0\} }\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)+
$$
$$
\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))=0\}}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=0.
$$
In the following I will prove that the product of three measures is associative. This can easily be generalized to a finite number of measures and possibly to countably many measures. As the OP indicates in his/her post scriptum, it all boils down to proving the associativity result for the $\sigma$-algebras involved.
Let $(X_i,\Sigma_i, \mu_i)$ be a measure space, $i = 1,\ 2,\ 3$. In the following I will use the (standard) notations $$\Sigma_i \times \Sigma_j = \{A \times B : A \in \Sigma_i,\ B \in \Sigma_j\}$$ $$\Sigma_i \otimes \Sigma_j = \sigma\big(\Sigma_i \times \Sigma_j\big),$$ where $\sigma(\cdot)$ indicates the $\sigma$-algebra generated by the argument, i.e. the smallest $\sigma$-algebra that contains the argument. We can finally state the result we want to prove
CLAIM: $\Sigma_1 \otimes \Sigma_2 \otimes \Sigma_3 := \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) = \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) = \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$
Synopsis of the proof: $\Sigma_i \times \Sigma_j \subset \Sigma_i \otimes \Sigma_j$ is a $\pi$-system. Define the "good set" and apply the $\pi-\Lambda$-system Theorem. Then use the minimality of $\sigma(\cdot)$ several times.
PROOF: Notice that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 = \{A \times B : A \in \Sigma_1\otimes\Sigma_2,\ B \in \Sigma_3\},$$
then fix $B \in \Sigma_3$ and let $$\Lambda = \{A : A \in \Sigma_1 \otimes \Sigma_2,\ A \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)\}.$$
Clearly $\Sigma_1 \times \Sigma_2 \subset \Lambda$. Let's prove that $\Lambda$ is a $\Lambda$-system.
- $X_1 \times X_2 \times B \in \Sigma_1 \times \Sigma_2 \times \Sigma_3$ then $X_1 \times X_2 \in \Lambda$.
- Let $A_1,A_2 \in \Lambda$, $A_1 \subset A_2$. We need to show that $A_2 \setminus A_1 \in \Lambda$, i.e. we need to show that $(A_2 \setminus A_1) \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$. This is easy since $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra, indeed $$(A_2 \setminus A_1) \times B = (A_2 \times B) \setminus (A_1 \times B) = (A_2 \times B) \cap (A_1 \times B)^c \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
- Let $\{A_i\}$ be an increasing sequence of sets in $\Lambda$. We need to show that $\cup A_i \in \Lambda$. As before, this easily follows from the fact that $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra. Let's write out the details! $$\Big(\bigcup_{i=1}^{\infty}A_i\Big) \times B = \bigcup_{i=1}^{\infty}(A_i \times B) \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
We can finally apply the $\pi-\Lambda$-system Theorem to conclude that $\sigma (\Sigma_1 \times \Sigma_2) \subset \Lambda$ and hence $\sigma(\Sigma_1 \times \Sigma_2) \times B \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$
Since this is true for every $B \in \Sigma_3$ we obviously get that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3),$$ which in turn yields $$\sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
Notice we can apply the same reasoning to show that $$\sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
The other inclusion is a lot simpler: $\Sigma_1 \times \Sigma_2 \times \Sigma_3 \subset (\Sigma_1 \otimes \Sigma_2) \times \Sigma_3$, so that $$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3),$$ and similarly
$$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$$
This proves the claim. $\blacksquare$
What happens to the measures was covered by David C. Ullrich in the comments section above.
Best Answer
Assume that $A \in M_1 \otimes M_2$ such that
$\mu_1\times \mu_2(A)=0$. We have to show that $\nu_1\times \nu_2(A)=0$.
By Fubini theorem we have
$$0=\mu_1\times \mu_2(A)=\int_{X_1}\mu_2(A \cap (x \times X_2))d\mu_1(x).$$
This means that $$ \mu_{1}(\{ x : \mu_2(A \cap (x \times X_2))>0 \})=0, $$
equivalently
$$ \mu_{1}( X_1 \setminus \{ x : \mu_2(A \cap (x \times X_2))=0 \})=0. $$
Since $\nu_2 \ll \mu_2$ we have
$$ \{ x : \mu_2(A \cap (x \times X_2))=0 \} \subseteq \{ x : \nu_2(A \cap (x \times X_2))=0 \}. $$
Since $\nu_1 \ll \mu_1$ and
$$ \mu_{1}( X_1 \setminus \{ x : \nu_2(A \cap (x \times X_2))=0\})=0, $$ we have $$ \nu_{1}( X_1 \setminus \{ x : \nu_2(A \cap (x \times X_2))=0\})=\nu_{1}(\{ x : \nu_2(A \cap (x \times X_2))>0\})=0. $$ Finally, we get
$$ \nu_1\times \nu_2(A)=\int_{X_1}\nu_2(A \cap \{x \times X_2\})d\nu_1(x)=$$
$$\int_{\{ x : \nu_2(A \cap (x \times X_2))>0\} }\nu_2(A \cap (x \times X_2))d\nu_1(x)+$$
$$\int_{\{ x : \nu_2(A \cap (x \times X_2))=0\}}\nu_2(A \cap (x \times X_2))d\nu_1(x)=0. $$