I don't know exactly what motivated the definitions of e and the natural logarithm, but I can provide a possible development that my analysis class did. It defines the natural logarithm and uses it to define e. This might not be what you are looking for, but it always fascinated me to see a way to develop logarithms from only some small properties.
Let's say we want to make a function L(x) with the property that L(xy) = L(x) + L(y) for all x and y in the domain. Let's investigate its properties and then form a definition to encompass those properties.
First of all, if 0 is in the domain of L(x), then we'd have L(x*0) = L(0) = L(0) + L(x) and L(x) = 0 for all x. Thus, to make a more interesting function, let's forbid 0 from being in the domain.
Now, we see that L(x*1) = L(x) = L(x) + L(1), so L(1) = 0.
Furthermore, L(-1*-1) = L(1) = L(-1) + L(-1), so L(-1) = 0 as well.
In fact, for any x in the domain of L,
L(x*x) = L(x^2) = L(-x*-x) = 2L(x) = 2L(-x), so L(x) = L(-x) and the function L is an even function.
Now, let's consider a L(x) which is differentiable and examine its properties.
We know that for any fixed y, we can differentiate with respect to x to get ' = ' = L'(x) = yL'(xy), where the second part comes from the Chain Rule. By now substituting x = 1, we get L'(1) = yL'(y), or L'(y) = L'(1) / y. Since L(x) is an even function, it has an odd derivative, so L'(y) = L'(1) / y holds for all y which are not 0.
Let K = L'(1). We know from the Fundamental Theora of Calculus that the integral from 1 to y of Kdx / x is L(y) - L(1) = L(y), if y > 0.
So for positive y, we have L(y) = K dx / x.
If y < 0, then L(y) = L(|y|) = K dx / x, since L is an even function. This is the most general rule which describes ALL logarithms. If we let K=1, and we restrict our domain to positive numbers, then we get the natural logarithm function. However, note that if K = 1 / ln b for some positive b, then we get logarithms to the base b. It's interesting to see that, as a consequence of this reason, the function L(x) = K dt / t is the MOST GENERAL FUNCTION that fulfills L(xy) = L(x) + L(y). Hence, the only functions which do this are logarithms!
e is defined as the unique number > 0 such that L(e) = 1. This is unique, because L(x) is strictly increasing for x > 0. Once you develop Taylor Polynomial approximations, then you can prove that
0> (1 + ax)^(1/ax) = e^a, and this justifies the formula that most people know for e.