Please prove the following: Given $ƒ(x) = e^x$, verify that $\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$. Given $ƒ(x) = e^x$, verify that 
$$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$  and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.
A proper answer to this question is fully dependent on the proper definition of symbol $e^{x}$. And further a proper definition of $e^{x}$ can not be given by just defining $e$.

One definition of $e^{x}$ is given by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ Although it is normally self-evident because of the symbol $n$, I want to emphasize that the $n$ in the above definition is a positive integer. Based on this definition it can be proven that $$e^{x + y} = e^{x}\cdot e^{y}\tag{2}$$ and $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{3}$$ Both the equations $(2)$ and $(3)$ are proved in this answer based on definition $(1)$. Using these equations $(2), (3)$ it is easy to prove that $$\frac{d}{dx}\,e^{x} = e^{x}\tag{4}$$ Second part of the question requires us to define the symbol $\ln$ (or $\log$ properly). One such definition is that if $e^{y} = x$ then $y = \ln x$ and then $\ln e = 1$ and this solves the second part of your question.
Using rules of exponentials,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = \lim_{h \rightarrow 0} \frac{e^xe^h-e^x}{h}$$
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \rightarrow 0} \frac{e^h-1}{h}$$

This has a limit of $1$ (although I'm not sure how to show this without using L'Hospital's rule. Therefore,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x$$