I think it is a good thing to talk about how there are some concepts where there are choices for where you start when definining them. It happens in linear algebra too, with the definition of linear dependence. You need to talk about how there is this web of connected properties, and it depends on what you're trying to achieve as to where you start. I do recommend prefacing any discussion of it with a general discussion on logs and exponentials using familar numbers, so they get a feel for how they ought to work first. This can even be done without graphs.
As to which definition of e I prefer...
Firstly, it's the functions ln(x) and $e^x$ that are the important thing, rather than the number $e$ itself per se. The number $e$ just crops up as a useful number to unify things.
I quite like the one where ln(x) is defined to be $\int_1^x \frac{1}{t} dt$, and e is the number for x that makes this area come out to 1. I have several reasons:
There is a real need for this function to have an integral because no power of $x$ differentiates to give $\frac{1}{x}$.
You can approximate e by using upper and lower sums of an easy function like 1/x.
The area truly exists because you can draw it there on the page, whereas it's a bit dicey as to whether it's reasonable to differentiate a function like $a^x$ when you don't know what it even means when $x$ is irrational!
Defining e using the area of a shape we can draw makes it feel a bit like $\pi$, which can be defined as the area of a circle and approximated by polygonal areas, and is one of the other famous irrationals we are familar with.
EDIT: For reference, here is the full sequence of definitions as I understand it (since it has been hotly debated in the comments):
As far as I understood it, it goes like this: for positive real $a$ you still define $a^1:=a$, $a^n:=a^{n−1}\times a$ for $n\in \mathbb{N}$, $a^0:=1$, $a^{−n}:=\frac{1}{a^n}$ for $n\in\mathbb{N}$, $a^\frac{1}{n}:=\sqrt[n]{a}$ for $n\in\mathbb{N}$ and $a^\frac{m}{n}:=(a^\frac{1}{n})^m$ for $n\in\mathbb{N}$ and $m\in\mathbb{Z}$.
Then you can still prove all the familiar properties of powers from these definitions, and that $\frac{d}{dx}x^r=rx^{r−1}$ for any rational $r$ using methods of first principles, without using logs at all. The issue is how you define $a^x$ for irrational $x$, because it needs a definition before you can differentiate $a^x$ or indeed define an inverse for the function.
If we want, can implicitly assume it does have a consistent definition which works for all the other properties we observe about powers, and investigate the behaviour of this function and its inverse and their derivatives, calling the inverse $\log_a (x)$, but realising that there is a hole in our definition of it and a hole in our arguments about derivatives until we know what $a^x$ means for irrational $x$.
Now we define $\ln(x) := \int_1^x \frac{1}{t}dt$ which immediately gives us that $\frac{d}{dx} \ln(x) = \frac{1}{x}$. From this definition we are able to prove all the familiar log laws for $\ln$. For example, to prove that $\ln(x^r) = r\ln(x)$ we can do the following:
By the chain rule, $\frac{d}{dx} \ln(x^r) = \frac{1}{x^r}\cdot rx^{r-1} = \frac{r}{x}$. However $\frac{d}{dx} r\ln(x) = \frac{r}{x}$. Since these two functions have the same derivative, they differ by a constant and so $\ln(x^r) = r\ln(x) +C$ for some $C$. Substituting $x=1$ gives $\ln(1) = r\ln(1) + C$, so $C=0$ and $\ln(x^r) = r\ln(x)$. Note that this only works for $r$ rational, because we still don't know what $x^r$ means for $r$ irrational.
We define $\exp$ to be the inverse of $\ln$, and using the properties of $\ln$ we are able to prove that it behaves the same as the powers laws we know. For example, we can prove that $\exp(a+b) = \exp(a)\times\exp(b)$ and $\exp(rx) = \exp(x)^r$. However this last one is only currently proved for rational $r$.
We define the particular number $e$ as the number such that $ln(e)=1$ or equivalently $e := exp(1)$, because we notice that $\exp(x)$ is the same as $e^x$ for all rational $x$ (and we know what $e^x$ is for rational $x$ because we defined what rational powers meant already without reference to $\exp$ or $\ln$). So now we define $e^x := \exp(x)$ for irrational $x$ and we define $a^x := \exp(\ln(a)\times x)$ for irrational $x$. We don't need to define $a^x$ for rational $x$ because it already has a definition, but we notice that even for rational $x$ this new calculation gives the correct result. We also notice that now $\exp(x) = e^x$ for all real $x$.
Finally, we can prove how derivatives of power functions work, and we can define $\log_a$ as the inverse of the function sending $x$ to $a^x$. And finally we observe that $\log_e(x)$ is the same as $\ln(x)$ for all real $x$.
This seems like a roundabout way of doing things, but if we define $a^x$ for $x$ irrational as the limit of a sequence, then all our proofs have to involve how limits of sequences interact with these functions and their derivatives and they are therefore actually harder.