I'm working through "The Calculus Tutoring Book" by Carol and Robert Ash (0-7803-1044-6). In chapter 3.3, when discussing derivatives of basic functions, they show and define the $D_{x}e^{x}$ and subsequently define $e$. Page 67 for those that have the book.
In any event, they define a "base", eventually, they land at $(1)$: $$D_{x}b^x = \lim_{\Delta x\to 0} b^x\lbrack \frac{b^{\Delta x}-1}{\Delta x} \rbrack$$ I have no problem here. And I understand why they effectively ignore the constant $b^x$ when you factor it out $(2)$: $$\frac{b^{\Delta x}-1}{\Delta x}$$ to $(3)$ $$D_{x}b^x=mb^x$$
And go on to state that the latter must be the slope of a line centered at $(0, 1)$
Then from the former, they land on the fact the $(4)$ $$D_{x}e^x = e^x$$
I don't see the path from $(2)$ to $(3)$ and subsequently $(4)$.
What they are doing, which is fairly sloppy the way it is phrased and justified, is the following.
They assume that limits as $\Delta x\to0$ of expressions of the form $(2)$ exist, and that they are numbers $m(b)$ (depending on $b$)
They argue "by picture" that, as $b$ moves between $1$ and $100$, the numbers $m(b)$ range from close to zero to very big
They assume that $m(b)$ depends continually on $b$
They assume that a continuous function satisfies the Intermediate Value Theorem
They put together all the above to conclude that there exists a number $e$ such that $m(e)=1$.
It follows, by $(1)$, that $(e^x)'=e^x$.
Comment: the notation you are using for derivatives is highly unusual, and it will be confusing for anybody who sees is. One often writes $f'(x)$ for the derivative of the function $f$ at the point $x$. When you write $f'(e^x)$, what we all read is "the derivative of the function $f$, evaluated at the point $e^x$", which is not what you meant.