To state the question more clearly, we are interested in
$$
\#\{ n\in\Bbb N,\, n\text{ odd}\colon n/2^{\Omega(n)} \le x\}
$$
where $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity. There is probably a standard method for attacking this type of problem directly (compare for example to classical results for $\#\{n\colon \phi(n)\le x\}$). But to orient ourselves, we can proceed by writing
$$
\#\{ n\in\Bbb N,\, n\text{ odd}\colon n/2^{\Omega(n)} \le x\} = 1 + \sum_{w=1}^\infty \#\{ n\in\Bbb N,\, n\text{ odd}\colon \Omega(n) = w,\, n \le 2^wx \}.
$$
Without the restriction to $n$ odd, this simply references the classical Selberg–Sathe result
$$
\#\{ n\le y\colon \Omega(n)=w\} \sim G\biggl( \frac{w-1}{\log\log y} \biggr) \frac{y(\log\log y)^{w-1}}{(w-1)!\log y}
$$
where
$$
G(z) = \frac1{\Gamma(z+1)} \prod_p \biggl( 1-\frac zp\biggr)^{-1} \biggl( 1-\frac1p \biggr)^z.
$$
We would then get a sum like
$$
\sum_{w=1}^\infty G\biggl( \frac{w-1}{\log\log y} \biggr) \frac{2^wx(\log\log x)^{w-1}}{(w-1)!\log x}
$$
(in practice $\log(2^wx)$ is close enough to $\log x$) which could be analyzed by standard means; indeed, the maximal summand occurs near $w\approx 2\log\log x$, in a region where the $G()$ value is basically constant, and we'd simply get the series for $\approx G(2) \frac x{\log x} e^{2\log\log x} \approx G(2) x\log x$.
I recommend going to a treatment of the Selberg–Sathe result (Montgomery & Vaughan Section 7.4, or Tenenbaum Chapter 2 Section 6.1) and modifying the method to restrict to odd numbers. One should be able to get an asymptotic formula of size $x\log x$, with the right leading constant even. [Technical note: the Selberg–Sathe result only holds as stated uniformly for $w\le(2-\delta)\log\log x$, which is just short of what we need. However, when restricted to odd integers, the range of uniformity will extend to $w\le(3-\delta)\log\log x$.]
