How can I prove that there is a function that is its own derivative? How can I prove that a function that is its own derivative exists? And how can I prove that this function is of the form $a(b^x)$?

There are two ways you could show it. The harder route would be to prove the existence and uniqueness theorem for ordinary differential equations, thus showing there exists solutions to $y'=y$. 
The more direct way would be to just construct the function $e^x$ and show that it's its own derivative. You would start by defining
$$\ln(x) = \int_1^x \frac{1}{t}\, dt$$
and prove that it's a strictly increasing function on $(0,\infty)$ with range $(-\infty, \infty)$. It follows that $\ln(x)$ has an inverse, which we should dub $e^x$. As for finding the derivative of this new and mysterious function: 
$$y=e^x$$
$$\ln(y)=x$$
Taking the $x$ derivative of both sides,
$$\frac{y'}{y} = 1$$
$$\implies y'=y$$
And do show that every function which is its own derivative is a constant multiple of $e^x$, suppose that $f'=f$. Then, noting that $e^x$ is nowhere zero,
$$\frac{d}{dx} \frac{f(x)}{e^x} = \frac{f'(x)e^x-f(x)e^x}{(e^x)^2} = \frac{f(x)e^x-f(x)e^x}{(e^x)^2} = 0$$
Therefore, $$\frac{f(x)}{e^x}$$ is constant since it has a connected domain, and so $f(x) = ce^x$ for some $c$. 
