How unique is $e$?

Assume that $f(x)$ is a function such that $f'(x)=f(x)$ for all $x\in\Bbb{R}$. Consider the quotient $g(x)=f(x)/e^x$. We can differentiate $$ g'(x)=\frac{f'(x)e^x-f(x)D e^x}{(e^x)^2}=\frac{f(x)e^x-f(x)e^x}{(e^x)^2}=0. $$ By the mean value theorem it follows that $g(x)$ is a constant. QED.

Consider the equation $y'=y$. Our goal is to solve for the function $y=f(x)$. Roughly speaking $$\frac{dy}{dx}=y \implies \frac{dy}{y}=dx \implies \int\frac{dy}{y}=\int dx \implies\ln(y)=x+C \implies y=e^{x+C}=Ae^x$$

for some constant $A$

This may not be an answer you are looking for, but its a nice one to consider.

Consider $y=\cos(ix)-i\sin(ix)$.

You may find that:

$$\frac{dy}{dx}=-i\sin(ix)-i^2\cos(ix)=\cos(ix)-i\sin(ix)$$

Thus, $y'=y$ is satisfied. Since $y(0)=1$, $y'(0)=1$, $\dots$, then by Taylor's theorem, we have $e^x=\cos(ix)-i\sin(ix)$, or more commonly known as

$$e^{ix}=\cos(x)+i\sin(x)$$

Which is Euler's formula for complex exponents.