real analysis - Intuitive explanation of $y' = y implies y = Ce^x$

I understand why $f : \mathbb{R} \to \mathbb{R}$ with $f'(x) = f(x)$ and $f(0) = 1$ must be $f (x) = e^x$, but I don't really feel it is super intuitive. Intuitively, why would you expect such a function to satisfy

$$f(a)f(b) = f(a+b)$$ or have exponential growth?

To build intuition, I tried the discrete case first, i.e.,

$$\frac{f(x+h) - f(x)}{h} = f(x) \implies f(x+h) = f(x)(h+1)$$

so

$$f(y) = f(0) (1+h)^{\frac{y}{h}} = f(0) c_h^{y}$$

where $c_h = (1+h)^{\frac{1}{h}}$ and saw what happens when $h \to 0$. However, I'm interested in a more intuitive explanation, if there's one. Bonus if it also explains $y' = P'y \Rightarrow y = Ce^P$ in a nice intuitive way.