There are infinitely many differentiable functions $F\colon [0,\infty)\to\mathbb{R}$ that satisfy
$$
F'(x)=F(2x)\qquad\text{and}\qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = F\left(\dfrac{2^x}{\log 2}\right)$. Plugging this into the above equation gives
$$
G\,'(x) = 2^x G(x+1)\qquad\text{and}\qquad \lim_{x\to-\infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G\,'(x-1)\tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - \int_{x+1}^1 2^{u-1} G(u)\,du\tag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^\infty$ function on $[0,1]$.
If we start with an arbitrary $C^\infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^\infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G\,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) \;=\; \sum_{i=0}^n \binom{n}{i}(-\log 2)^{n-i}\,2^{-x}\, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) \;=\; \sum_{i=0}^n \binom{n}{i}(-\log 2)^{n-i}\, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^\infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0\;\;\text{ for all }n\geq 1
\qquad\text{and}\qquad
G(1)=0.
$$
The Limit as $x\to-\infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - \int_{x+1}^1 2^{u-1} G(u)\,du\tag*{(2)}
$$
If we assume that $G(x)\geq 0$ for $x\in [0,1]$ and, say, that
$$
\int_0^1 2^{u-1} G(u)\,du \;<\; \frac{G(0)}{10}
$$
then $\frac{9}{10}G(0) \leq G(x) \leq G(0)$ for all $x\in[-1,0]$. It follows recursively that
$$
G(0) \;\geq\; G(x) \;\geq\; \left(\frac{9}{10} - \frac{1}{\log 4}\right)G(0) \;>\; 0
$$
for all $x<-1$, since
\begin{align*}
G(x) \;&=\; G(0)-\int_{x+1}^1 2^{u-1}G(u)\,du \\
&=\; G(0) - \int_0^1 2^{u-1}G(u)\,du - \int_{x+1}^0 2^{u-1}G(u)\,du \\
&\geq\; G(0) - \frac{G(0)}{10} - \int_{-\infty}^0 2^{u-1}G(0)\,du \\
&=\; \left(\frac{9}{10} - \frac{1}{\log 4}\right)G(0).
\end{align*}
Then $G$ is monotone on $(-\infty,0)$, so $\displaystyle\lim_{x\to-\infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
