Short answer: Yes to your questions.
Your reasoning is intuitively correct. As pointed out, the event that two given primes divides $N$ are roughly independent, so the number of prime factors roughly follows a Poisson distribution.
The trick is to go from "approximately" to some precise estimate. We can make it rigorous by showing that as $y\to\infty$, the distribution of the number of divisors in $[2^{x-1}, 2^x-1]$ of $N$ uniform in $[2^{y-1}, 2^y-1]$ converges to an explicit distribution (that of a sum of independent random variables, hence justifying your initial, intuitive approximation). This sum of independent variables can then be controlled as $x\to\infty$, giving your asymptotics in $\frac 1 x$. Note the double limit, which makes it delicate to quantify the quality of the approximation.
Step 1: number of integers in $[A,B-1]$ without divisors in $Q \subset P$.
Let $P$ be the set of prime numbers, and $Q\subset P$. For every integer $m$, there are exactly $\left\lfloor \frac{n}{m}\right\rfloor$ integers in $\{1, \dots, n\}$ that are divisible by $m$. By the inclusion-exclusion principle, the number of integers in $\{1, \dots, n\}$ that are divisible by no $q\in Q$ is
$$ E_Q([1,n]) := \sum_{J\subset Q} (-1)^{|J|} \left\lfloor \frac{n}{\prod_{q\in J} q} \right\rfloor , $$
and for $A < B$, the set of integers in $[A, B-1]$ that are divisible by no $q\in Q$ is
$$ E_Q([A, B-1]) := \sum_{J\subset Q} (-1)^{|J|} \left( \left\lfloor \frac{B-1}{\prod_{q\in J} q} \right\rfloor - \left\lfloor \frac{A-1}{\prod_{q\in J} q} \right\rfloor \right) . $$
Observe that since $|x-\lfloor x\rfloor|\leq 1$,
$$ \left| E_Q([A,B-1]) - (B-A) \sum_{J\subset Q} \frac{(-1)^{|J|}}{\prod_{q\in J} q} \right| \leq 2 \sum_{J\subset Q} 1 = 2^{1+|Q|} . $$
We can simplify this further by noticing that
$$ \sum_{J\subset Q} \frac{(-1)^{|J|}}{\prod_{q\in J} q} = \prod_{q\in Q} \left(1 - \frac 1 q \right) . $$
Step 2: the moment generating function.
Let $x < y$ be integers, and for every $n$, let $D(n)$ be the number of prime divisors of $n$ in $[2^{x-1}, 2^x-1]$. Let $N$ be a uniformly distributed integer in $[2^{y-1}, 2^y-1]$. We have
$$ \mathbb E[z^{D(N)}] = 2^{1-y} \sum_{n=2^{y-1}}^{2^y-1} z^{D(n)} . $$
The latter can be rewritten as follows. Let $Q = [2^{x-1}, 2^x-1] \cap P$. Then by partitioning the $n$ depending on the set $I$ of $p\in Q$ that divide it, since $n / \prod_{p \in I} p$ is divided by no $p \in Q \setminus I$ we get
$$ \mathbb E[z^{D(N)}] = 2^{1-y} \sum_{I\subset Q} z^{|I|} E_{Q\setminus I} \left( \left[ \left\lfloor\frac{2^{y-1}}{\prod_{p\in I} p}\right\rfloor , \left\lfloor\frac{2^{y}-1}{\prod_{p\in I} p}\right\rfloor \right] \right) . $$
Using the approximation from step 1,
$$ (*) \quad \left| \mathbb E[z^{D(N)}] - \sum_{I\subset Q} \frac{z^{|I|}}{\prod_{p\in I} p} \prod_{q \in Q\setminus I} \left(1 - \frac 1 q \right) \right| \leq 2^{1-y} \sum_{I\subset Q} |z|^{|I|} 2^{1+|Q|-|I|} = 2^{2-y} (2 + |z|)^{|Q|} . $$
We can identify $\sum_{I\subset Q} \frac{z^{|I|}}{\prod_{p\in I} p} \prod_{q \in Q\setminus I} \left(1 - \frac 1 q \right) $ with $\mathbb E[z^{\sum_{p\in Q} B_p}]$, where $(B_p)_{p\in P}$ are independent random variables with $P(B_p = 1) = 1/p$ and $P(B_p = 0) = 1-1/p$. Since the right-hand side of $(*)$ converges to $0$ as $y\to\infty$, we deduce the following Lemma.
Lemma: Recall that $N$ is a uniform integer in $[2^{y-1}, 2^y-1]$, that $Q$ is the set of prime numbers in $[2^{x-1}, 2^x-1]$, and that $D(N)$ is the number of distinct prime divisors of $N$ in $Q$. As $y\to\infty$, $D(N)$ converges in distribution towards $\sum_{p\in Q} B_p$, where $(B_p)_{p\in P}$ are independent random variables with $P(B_p = 1) = 1/p$ and $P(B_p = 0) = 1-1/p$. In particular, for every fixed $k$, $P(D(N) \geq k) \to P(\sum_{p\in Q} B_p \geq k)$ as $y\to\infty$.
Step 3: Asymptotic as $x\to\infty$.
Immediately
$$ P(\sum_{p\in Q} B_p = 0) = \prod_{p\in Q} \left(1 - \frac 1 p\right) . $$
There is a small tricky thing here: Merten's theorem, in its weaker form, only gives you $\sum_{p\in Q} \frac 1 p = O\left(\frac 1 x\right)$; you need one of the stronger versions to get that $\sum_{p\in Q} \frac 1 p = \frac 1 x + o\left(\frac 1 x\right)$. Still, you can get it, and it gives you the desired asymptotic $ x P(\sum_{p\in Q} B_p \geq 1) \to 1$ at $x \to\infty$.
Since
$$ P(\sum_{p\in Q} B_p = 2) = \prod_{p\in Q} \left(1 - \frac 1 p\right) \sum_{p < p \in Q} \frac{1}{(p-1)(p'-1)} \approx \frac 1 2 \left(\sum_{p\in Q} \frac 1 p\right)^2 \approx \frac 1 {2x^2}$$
your second asymptotic follows once you are convinced that $P(\sum_{p\in Q} B_p \geq 3)$ is much smaller than $P(\sum_{p\in Q} B_p = 2)$. This, and the approximation in the last line, can be made rigorous.
