A very weird integral equation

You can assume that $f$ is a power series.

$$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots = \sum_{n=0}^{\infty} a_n x^n$$

Then

$$\sum_{n=0}^{\infty} a_n x^n = \int_x^{2x} \left( \sum_{n=0}^{\infty} a_n t^n \right) dt$$

or that

$$a_0 + a_1 x + a_2 x^2 + \dots = a_0x + a_1 \frac{3}{2}x^2 + a_2 \frac{7}{3}x^3 + \dots$$

The LHS has a constant term $a_0$, but the RHS has no constant term. So $a_0 = 0$. But the $x^1$ coefficient $a_1$ must be $0$ since $a_1x = a_0x = 0x$. The logic continues, and every coefficient $a_n$ must be zero. We have that $f(x) = 0$.

It seems like it is illegal, but if you define $T$ as the operator on $\mathcal{C}^{1}$, representing $$T[f]=\left(\dfrac{d}{dx}-\text{id}\right)f$$ Then dividing $1-\dfrac{d}{dx}$ is same as taking the inverse operator of $T$ on both sides, in particular, write $g(x)=-2f(2x)$, then $$T[f](x)=g(x)\implies (T^{-1}\circ T)[f](x)=T^{-1}[g](x).$$ But on $\mathcal{C}^{1}$, this operator likely does not admit an inverse as it is not injective, so you do it on the quotient space $\mathcal{C}^1/\sim$ where $f\sim g\iff f(x)=g(x)+Ce^x$ for some constant $C$.

That's why the function you solved out is generally an equivalent class just like you solve indefinite integral, you need to add $+C$ at the end in this case you should add $+ Ce^x$.

But finally, when you differentiate the integral equation, you lose some information as we cannot distinguish it come from which element in the equivalence class. So the original integral equation is actually giving you more information than $f'(x)+2f(2x)-f(x)=0$. And precisely it gives $f(0)=0$. Thus using this initial condition you can show that the above $C$ is actually $0$.

Finally from the functional equation, since the above process we explicitly assume $f$ is differentiable, $f$ is continuous, so for every $y\in\mathbb{R}$, $$f(y)=\dfrac{1}{2}f\left(\dfrac{y}{2}\right)=\cdots=\dfrac{1}{2^n}f\left(\dfrac{y}{2^n}\right)\quad\forall~n\in\mathbb{N}.$$ So continuity of $f$ forces $f(y)\equiv f(0)$ for every $y$.

If the function only has to be defined on $[0,\infty)$, and if it doesn't have to be analytic, then there are nonzero solutions.

Consider the lacunary series $$f_q(x)\stackrel{\triangle}{=}\frac{1}{\prod_{k=1}^{\infty}(1-q^k)}\sum_{n=0}^{\infty}\frac{(-1)^n q^{\,_nC_2}}{\prod_{k=1}^n(1-q^k)}\mathrm{e}^{-q^{-n}x}$$ where $0<q<1$. It's uniformly convergent on $\Re x \geq 0$, so one can safely exchange summation and limiting operations to show that it's smooth and satisfies $$\begin{align}f_q(x) &= \int_x^{x/q}f_q(x')\mathrm{d}x'\\ \lim_{x\to\infty}\mathrm{e}^xf_q(x) &=\frac{1}{\prod_{k=1}^{\infty}(1-q^k)} \\ f_q^{(m)}(0^+) &= (-1)^m\sum_{n=0}^{\infty}\frac{(-1)^n q^{\,_nC_2-mn}}{\prod_{k=1}^n(1-q^k)} \end{align}$$ But this summation for $f_q^{(m)}(0^+)$ vanishes for all nonnegative integers $m$ by a $q$-binomial theorem, so $f_q$ cannot be analytic.

More generally, the Laplace transform of the integral equation $$f(x)=\int_x^{x/q}f(x')\mathrm{d}x'$$ is $$\hat{f}(\kappa) = \frac{\hat{f}(q\kappa)-\hat{f}(\kappa)}{\kappa}\text{.}$$ Substituting $\hat{f}(\kappa) = \hat{f}_q(\kappa)\hat{\psi}(\kappa)$ where $\hat{f}_q(\kappa)=\prod_{n=0}^{\infty}(1+q^n\kappa)^{-1}$ is the Laplace transform of $f_q(x)$ above yields the constraint $$\hat{\psi}(q\kappa)=\hat{\psi}(\kappa)\text{.}$$ That is, $f$ satisfying the integral equation are convolutions of $f_q$ with $q$-multiplicatively periodic distributions $\psi$.