With no other context than:
$$\frac{dy}{dx}=ay.$$
I understand the more popular solution that doesn’t involve imaginary numbers would simply be $y(x)=e^{ax}$.
But because of Euler’s identity, can I claim that $y(x)=\cos(x)+i\sin(x)$ is a general solution to the differential equation as well because $y’(x)=iy(x)$?
I am asking because if the differential equation is presented without claiming that $a\in\mathbb{R}$, can I just assume that $a$ is ‘allowed’ to take on imaginary values?
On
$y=\cos(x)+i\sin(x)\quad$ is NOT general solution of $\quad\frac{dy}{dx}=ay.$
$y=\cos(x)+i\sin(x)\quad$ is NOT particular solution of $\quad\frac{dy}{dx}=ay.$
If you put $\quad y=\cos(x)+i\sin(x)\quad$ into $\quad\frac{dy}{dx}=ay \quad$ you get $\quad -\sin(x)+i\cos(x)=a(\cos(x)+i\sin(x))\quad$ which obviously is false, even with $a=1$.
Even with $a=1$, you see that your claim $y'(x)=i\:y(x)$ is false with $y(x)=\cos(x)+i\:\sin(x)$.
What is correct ?
The general solution of $\quad\frac{dy}{dx}=ay \quad$ is : $\quad y=c\:e^{ax}\quad$ where $c$ is an arbitrary constant.
If you put $\quad y=c\:e^{ax}\quad$ into $\quad\frac{dy}{dx}=ay \quad$ you get $\quad c\:(a\:e^{ax})=a\:(c\:e^{ax})\quad$ which is correct.
Note that $\quad y=e^{ax}\quad$ is not the general solution. It is a particular solution which corresponds to the particular value of $c=1$.
Coming back to your question about the complex cases :
In the case where the parameter $a$ is complex, say $\quad a=r_1+i\:r_2\quad$ with $r_1$ and $r_2$ real, the general solution of $\quad \frac{dy}{dx}=(r_1+i\,r_2)y\quad$ is : $$y=c\:e^{(r_1+i\,r_2)x}$$ $$y=c\:e^{r_1x}\:e^{i\,r_2x}$$ Because of Euler’s identity, it is the same as : $$y=c\:e^{r_1x}\left(\cos(r_2x)+i\:\sin(r_2x)\right)$$
Based on This page from WolframAlpha, I would agree that a solution to $$\frac{dy}{dx} = ay$$ is indeed $$y = \cos(x) + i\sin(x)$$ due to the fact that the alternate form of that equation turns out to be $$y = e^{ix}$$ Mind blown, eh?