The answer to (1) is certainly no; not every measure on a product space is a product measure, not even for a finite product. Consider for example the following: let $m$ be Lebesgue measure on $[0,1]$, let $F : [0,1] \to [0,1]^2$ be given by $F(x) = (x,x)$, and let $\mu = m \circ F^{-1}$ be the pushforward measure on the product $[0,1]^2$ (with its product $\sigma$-algebra of course). $\mu$ then is effectively 1-dimensional Lebesgue measure on the diagonal of $[0,1]$. Now $\mu$ cannot be a product measure. For suppose $\mu = \mu_1 \times \mu_2$. Let $0 < a < 1$; it's clear that we have $\mu((0,a)^2) = a > 0$ and $\mu((a,1)^2) = 1-a > 0$. Thus $\mu_i((0,a)) > 0$ and $\mu_i((a,1)) > 0$ for $i=1,2$. But $\mu((0,a) \times (a,1)) = \mu((a,1) \times (0,a)) = 0$, a contradiction.
This is best thought of in terms of probability theory: a probability measure $\mu$ on $\mathbb{R}^d$ gives the (joint) distribution of a random vector $(X_1, \dots, X_d)$. If $\pi_i : \mathbb{R}^d \to \mathbb{R}$, $i=1,\dots,d$ is the projection onto the $i$'th component, then $\mu_i = \mu \circ \pi_i^{-1}$ is the marginal distribution of $X_i$. But $\mu$ is a product measure iff $\{X_1, \dots, X_d\}$ are independent. In our example, $\mu$ is the joint distribution of $(U,U)$, where $U \sim U(0,1)$; obviously $U$ is not independent of itself.
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.
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Trivial example: $X = \{x\}$ has one point, $\mu(\{x\}) = \infty$, $X_1 = X_2 = X$, and $\mu_1 = \mu_2 = \mu$. Not $\sigma$-finite, but the unique product measure is $\lambda(\{(x,x)\}) = \infty$.