As I pointed out above, we need to remove the origin, so the space we're interested in is $R = \mathbb{C}^*/\sim$. Note that we can view $R$ as the quotient of a group action, namely $\mathbb{Z}\times\mathbb{C}^* \to \mathbb{C}^*$ where $(n, z) \mapsto 2^nz$.
It is not true that an open neighbourhood of $w \in \mathbb{C}^*$ necessarily has infinitely many equivalent points under $\sim$. For every $a \in (0, \infty)$, the half-open annulus $\{z \in \mathbb{C} \mid a \leq |z| < 2a\}$ is a fundamental domain of the group action, i.e. every point in $\mathbb{C}^*$ is equivalent to a unique point in the annulus. So for $w \in \mathbb{C}^*$, choose $a$ such that $\frac{1}{2}|w| < a < |w|$, then $a < |w| < 2a$ so $\{z \in \mathbb{C}^* \mid a < |z| < 2a\}$ is an open neighbourhood of $w$ in which no two points are equivalent.
Let $\pi : \mathbb{C}^* \to \mathbb{C}^*/\sim$ be the quotient map. We put a complex structure as follows: let $(\pi(U), \varphi)$ be charts where $V = \pi(U)$ for some $U$ with no two points equivalent and $\varphi : V \to U$ such that $\varphi_w(\pi(z)) = z$ - note, such a $\varphi$ exists because no two points of $U$ are equivalent so each point in $\pi(U)$ has a unique preimage. One can check that these charts are compatible and define a maximal atlas. Note that $R = \mathbb{C}^*/\sim\, = \pi(\mathbb{C}^*) = \pi(\{z \in \mathbb{C} \mid 1 \leq |z| \leq 2\})$. As $R$ is the continuous image of a compact set, it is itself compact.
It is not hard to see that, topologically, $R$ is a torus as it can be obtained by identifying the inner and outer boundaries of the closed annulus $\{z \in \mathbb{C} \mid 1 \leq |z| \leq 2\}$ - note, this is exactly what the quotient map $\pi$ does. A torus is usually constructed as a parallelogram with opposite sides identified. These two constructions really are the same because by identifying one pair of opposite sides of the parallelogram, we obtain a closed annulus where the unidentified sides become the inner and outer boundaries.
More generally, if $\alpha \in \mathbb{C}^*$ with $|\alpha| > 1$, and we define $\sim$ by $z \sim w$ if and only if $z = \alpha^nw$, then the above argument shows that $R_{\alpha} := \mathbb{C}^*/\sim$ is a compact Riemann surface which is topologically a torus. Therefore $R_{\alpha}$ is of the form $\mathbb{C}/\Lambda$ for some lattice $\Lambda$. We can choose the lattice to be of the form $\langle 1, \tau\rangle$ where $\tau \in \mathbb{H} = \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0\}$. There should be some correspondence between $\tau$ and $\alpha$, once we've determined what it is, we just need to find the $\tau$ corresponding to $\alpha = 2$.
We want to find a biholomorphism $\mathbb{C}/\Lambda \to R_{\alpha}$ for appropriate choices of $\tau$ and $\alpha$. One way to construct such a map is to begin with a holomorphic map $\mathbb{C} \to \mathbb{C}^*$ which descends to the quotients. A natural candidate is the exponential map.
Consider $f : \mathbb{C} \to \mathbb{C}^*$ given by $f(z) = \exp(-2\pi iz)$. Note that if $m + n\tau \in \Lambda$,
$$f(z + m + n\tau) = f(z)f(m)f(n\tau) = f(z)f(\tau)^n.$$
So if we take $\alpha = f(\tau)$, then $z$ and $z + m + n\tau$ have the same value under the induced map $\mathbb{C} \to R_{\alpha}$ given by $z \mapsto [f(z)]$. Therefore, it descends to a well-defined map $F : \mathbb{C}/\Lambda \to R_{\alpha}$ given by $z + \Lambda \mapsto [\exp(-2\pi iz)]$. In local coordinates, the map $F$ is given by $z \mapsto f(z - \lambda)\alpha^k$ for some $\lambda \in \Lambda$ and some $k \in \mathbb{Z}$. As $f$ is holomorphic, so too is $F$. Note that if $[f(z)] = [f(w)]$ then $f(z) = f(w)f(\tau)^n = f(w + n\tau)$ and therefore $a = w + n\tau + m$, so $z + \Lambda = w + \Lambda$. That is, $F$ is injective and therefore a biholomorphism between $\mathbb{C}/\Lambda$ where $\Lambda = \langle 1, \tau\rangle$ and $R_{\alpha}$ where $\alpha = f(\tau) = \exp(-2\pi i\tau)$.
As $f$ is surjective, for any $\alpha \in \mathbb{C}^*$, we can find an appropriate $\tau$. If $|\alpha| > 1$, then
$$1 < |\alpha| = |\exp(-2\pi i\tau)| = \exp(\operatorname{Re}(-2\pi i\tau)) = \exp(2\pi\operatorname{Im}(\tau)),$$
so $\tau \in \mathbb{H}$. In this particular case, we just need to find a $\tau$ satisfying $\exp(-2\pi i\tau) = 2$. One option is $\tau = \frac{i}{2\pi}\ln 2$.
I have attempted to indicate what $f$ does in the image below. Note that on the left, both pairs of sides (the green and the blue pairs) need to be identified to obtain a torus, while on the right, the green pair has already been identified, so only the blue pair remains.
Note, the green curve on the right from $1$ to $\tau$ can wind in the clockwise or anticlockwise direction.
Consider $\gamma : [0, 1] \to \mathbb{C}$ given by $\gamma(t) = t\tau$. This is a parameterisation of the green line between $0$ and $\tau$ in the left picture. Composing with the map $f$, we get a parameterisation of the green curve on the right. If $\tau = a+bi$, then
$$(f\circ\gamma)(t) = \exp(-2\pi it\tau) = e^{2\pi bt}(\cos(-2\pi at) + i\sin(-2\pi at)).$$
If $a > 0$, then $\sin(-2\pi at) < 0$ for small $t$, so the curve winds clockwise. If $a < 0$, then $\sin(-2\pi at) > 0$ for small $t$, so the curve winds anticlockwise. If $a = 0$, then $\sin(-2\pi at) = 0$ so $\tau$ lies on the real axis and $f\circ \tau$ is a parameterisation of the straight line segment from $1$ to $\tau$.
Moreover, if $a \neq 0$, the curve can also wind multiple times.
Each time the curve winds around the centre circle, it crosses the real axis twice. Therefore, if the solutions to $\operatorname{Re}((f\circ\gamma)(t)) = 0$ for $t \in [0, 1]$ are $0, t_1, \dots, t_k$, then the number of winds is $\left\lfloor\frac{k}{2}\right\rfloor$. Note that $\operatorname{Re}((f\circ\gamma)(t)) = e^{2\pi bt}\sin(-2\pi at) = 0$ is equivalent to the equation $\sin(-2\pi at) = 0$. The zeroes of sine are integer multiples of $\pi$, so we need to determine for how many integers $m$ is there $t \in [0, 1]$ such that $-2\pi at = m\pi$; equivalently, how many integers $m$ such that $t = -\frac{m}{2a} \in [0, 1]$. There are $\lfloor 2a\rfloor + 1$ such integers, so $k = \lfloor 2a\rfloor$ and hence the green curve winds $\left\lfloor\frac{\lfloor 2a\rfloor}{2}\right\rfloor = \lfloor a\rfloor$ times. Note, the sign of the winding indicates the orientation: positive winding is clockwise, while negative winding is counterclockwise.
