Uniqueness does not hold in general.
Besides the usual product measure of two measure spaces $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$, there is a different version of the product measure, called the complete locally determined product. For products of non-$\sigma$-finite measure spaces it has many desirable properties that the usual product measure lacks. In his Measure Theory, Fremlin goes so far as to call the usual product measure the primitive product measure, see chapter 5 of volume 2 for more details on the construction and its basic properties.
What follows below is taken more or less directly from Fremlin's exposition.
The construction of the complete locally determined version of the product measure is as follows: Let $\pi = \mu \times \nu$ be the usual product measure on the full $\sigma$-algebra $\mathfrak{P} = \mathfrak{M} \otimes \mathfrak{N}$ obtained from performing the Carathéodory construction on the measurable rectangles. It seems more natural to do this than to restrict to the $\sigma$-algebra $\mathfrak{M} \times \mathfrak{N} \subset \mathfrak{M} \otimes \mathfrak{N}$ generated by the measurable rectangles from the beginning. The complete locally determined product measure $p$ is an “inner regularization” of the usual product measure $\pi$. Its definition is that for $P \in \mathfrak{P}$ one puts
$$p(P) = \sup{\{\pi(P \cap (M \times N))\,:\,\mu(M) \lt \infty, \nu(N) \lt \infty\}}.$$
If you're willing to believe that $p$ is a measure, you can skip the next section of this answer.
Let us check that $p$ is indeed a measure: Clearly, $p(\emptyset) = 0$. If $P_n \in \mathfrak{P}$ is a sequence of pairwise disjoint sets, we can estimate for every pair $M \in \mathfrak{M},N \in \mathfrak{N}$ with $\mu(M) \lt \infty$ and $\nu(N) \lt \infty$ that
$$
\pi \left( \bigcup_{n=1}^{\infty} P_n \cap (M \times N)\right) = \sum_{n=1}^{\infty}\; \pi\left(P_n \cap (M \times N)\right) \leq \sum_{n=1}^{\infty}\;p(P_n),
$$
hence
$p\left(\bigcup_{n=1}^{\infty}P_n\right) \leq \sum_{n=1}^{\infty}\;p(P_n),$ so $p$ is $\sigma$-subadditive. To prove $\sigma$-additivity, note that we can choose for every $t \lt \sum p(P_n)$ a large enough $k$ such that there are
$t_1 \lt p(P_1), \ldots, t_k \lt p(P_k)$ with $t \lt t_1 + \cdots + t_k$. Then we can choose $M_i \in \mathfrak{M},N_i \in \mathfrak{N}$ with
$\mu(M_i)\lt \infty$ and $\nu(N_i) \lt \infty$ such that $t_i \leq \pi(P_i \cap (M_i \times N_i))$. Let $M = M_1 \cup \cdots \cup M_k$, $N = N_1 \cup \cdots \cup N_k$. Then, by definition, and the fact that $\mu(M) \lt \infty$ and $\nu(N)\lt \infty$, we have
$$
p\left(\bigcup_{n=1}^\infty P_n\right) \geq
\pi\left(\bigcup_{n=1}^\infty P_n \cap (M \times N)\right) \geq
\sum_{n = 1}^k \; \pi (P_n \cap (M \times N)) \geq
\sum_{n=1}^k \; \pi(P_n \cap (M_n \times N_n)) \gt t.
$$
Thus $t \lt p(\bigcup P_n) \leq \sum p(P_n)$ for every $t < \sum p(P_n)$. It follows that $p$ is $\sigma$-additive on disjoint sequences sets in $\mathfrak{P}$, hence $p$ is a measure on $\mathfrak{P}$.
Exercise 1: It is clear that $p(M \times N) = \mu(M)\,\nu(N)$ for $M$ and $N$ of finite measure. To ensure this equality for all measurable sets $M$ and $N$, we need to suppose that $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ are semi-finite in the sense that every set of infinite measure contains a measurable set of finite positive measure. This implies that $\mu(M) = \sup{\{\mu(E)\,:\,E \subset M, \mu(E) \lt \infty\}}$ and similarly for $\nu$. Using this, check that the equality $p(M \times N) = \mu(M)\, \nu(N)$ holds for all $M \in \mathfrak{M}$ and $N \in \mathfrak{N}$.
Exercise 2: Let $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ be $\sigma$-finite spaces. Prove that $p = \pi$.
Exercise 3: Let $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ be arbitrary measure spaces. If $\lambda$ is any measure on $\mathfrak{P}$ satisfying $\lambda(M \times N) = \mu(M)\nu(N)$ then $p(P) \leq \lambda(P) \leq \pi(P)$.
We are finally ready to concoct our counterexample—most of the basic counterexamples in connection with product measures are variations of this theme.
Let $(X,\mathfrak{M},\mu) = ([0,1],\Sigma,\lambda)$ be the unit interval with Lebesgue measure. Let $(Y,\mathfrak{N},\nu) = ([0,1],\mathcal{P}([0,1]),\#)$ be the unit interval equipped with counting measure. Let $\pi$ be the usual product measure and $p$ its locally determined version. Since both $\mu$ and $\nu$ are semi-finite, we have by exercise 1 above that $p(M \times N) = \mu(M) \nu(N)$ for all measurable $M \subset X$ and $N \subset Y$.
Notice that the diagonal $\Delta = \{(x,x) \,:\,x \in [0,1]\}$ is in the $\sigma$-algebra $\mathfrak{M \times N}$ generated by the measurable rectangles, because
$$\Delta = \bigcap_{n =1}^{\infty}\; \bigcup_{k=0}^{n-1}\; \left[\frac{k}{n},\frac{k+1}{n}\right] \times \left[\frac{k}{n},\frac{k+1}{n}\right].$$
Exercise 4: $\pi(\Delta) = \infty$.
Exercise 5: $p(\Delta) = 0$.
Thus, $\pi$ and $p$ are distinct, while agreeing on the measurable rectangles themselves. Since $\Delta \in \mathfrak{M} \times \mathfrak{N}$ this also holds for the restrictions of $\pi$ and $p$ from $\mathfrak{M \otimes N}$ to $\mathfrak{M \times N}$.
Best Answer
Partial answer: if $\nu$ is semifinite then it is $\sigma$-finite.
Following the setup of the first version of your statement, suppose that $\pi$ is $\sigma$-finite and satisfies $\pi(A \times B) = \mu(A) \nu(B)$ and that $\nu$ is semifinite. (We do not actually need the assumption that $\mu$ is $\sigma$-finite.) We show $\nu$ is $\sigma$-finite.
We follow the idea from Uniqueness of product measure (non $\sigma$-finite case). Since $\pi$ is $\sigma$-finite, there is a sequence of sets $E_n \subset X \times Y$ with $\pi(E_n) < \infty$ and $X \times Y = \bigcup_n E_n$. For each $n$, let $$a_n = \sup\{ \pi((X \times B) \cap E_n) : \nu(B) < \infty\}.$$ Note that $a_n \le \pi(E_n) < \infty$. Now by definition of sup, there is a sequence of sets $B_{n,k} \subset Y$ with $\nu(B_{n,k}) < \infty$ and $\pi((X \times B_{n,k}) \cap E_n) \uparrow a_n$. Set $C = Y \setminus \bigcup_{n,k} B_{n,k}$. I claim $\nu(C) =0$, which would complete the proof.
By semifiniteness, it is enough to show that for every measurable $C' \subset C$ with $\nu(C') < \infty$, we have $\nu(C') =0$. Now since such $C'$ is disjoint from all the $B_{n,k}$, we have for every $k$ that $$\pi((X \times B_{n,k}) \cap E_n) + \pi((X \times C') \cap E_n) = \pi((X \times (B_{n,k} \cup C')) \cap E_n) \le a_n$$ since $\nu(B_{n,k} \cup C') < \infty$. But by definition of the $B_{n,k}$, we have $\sup_k \pi((X \times B_{n,k}) \cap E_n) = a_n$, so we conclude $\pi((X \times C') \cap E_n) = 0$. Since $X \times Y$ is the countable union of the $E_n$, it follows that $0 = \pi(X \times C') = \mu(X) \nu(C')$. Since $\mu(X) > 0$ by assumption, we have $\nu(C') = 0$ as desired.
I don't know what happens if $\nu$ is not semifinite. You would think that if $\nu$ is not semifinite then it would make it even harder for $\pi$ to be $\sigma$-finite, but I don't see how to prove that at the moment.