I'm not looking for a technical proof. I'm quite familiar with things like Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ and how one could derive that by defining $e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}$ for $z\in\mathbb{C}$.
Rather, I'm looking for a rich intuitive understanding.
I get that the main interesting thing about exponentials is how they swap the roles of addition & multiplication. Which is to say, for $f:\mathbb{R}\rightarrow\mathbb{R}$, we could actually define the property of it being exponential by $f(a+b) = f(a)f(b)$ for all $a,b\in\mathbb{R}$ and then derive everything else we're familiar with about exponentials from that one property.
One could just replace $\mathbb{R}$ with $\mathbb{C}$ here and generalize nicely to exponential functions in the complex plane. And if $f(1)\in\mathbb{R}$, then $f|_\mathbb{R}$ will be just like a normal real exponential function by $f(\lambda) = (f(1))^\lambda$. Which is to say, $f$ extends a real exponential into the complex plane.
This has some nice geometry to it on the real number line. $\mathbb{R}$ stretches and shifts rightward such that $\mathbb{R}^-$ maps to $(0,1)$ and $1$ maps to the exponential base $f(1)$.
There are three points I'm not at all clear on:
Why would $f(i\mathbb{R})=S^1=\{z\in\mathbb{C} : |z|=1\}$? I see why it's plausible, since that gives $f(a+bi)=f(a)f(bi)$ for $a,b\in\mathbb{R}$ and that does in fact extend a real exponential into the complex plane if $|f(bi)|=1$. But why does it have to be that $|f(bi)|=1$? Geometrically speaking, why does a complex exponential have to wrap the imaginary line repeatedly around the unit circle?
I think that given some real exponential, there's exactly one extension to the complex plane. Which is to say, for a complex exponential $f$, the value of $f(1)\in\mathbb{R}$ is sufficient to completely define $f$. Is that true? If so, why?
Given the uncountably infinite number of exponential functions $f:\mathbb{C}\rightarrow\mathbb{C}^*$ with $f(1)\in\mathbb{R}$, why does the one that doesn't distort lengths in its mapping $f|_{i\mathbb{R}}: i\mathbb{R}\rightarrow S^1$ also happen to send $f(1)$ specifically to $e$? Why not some other real number?
This is an intuitive way of answering your third question.
Think of uniform circular motion around the unit circle in the complex plane, moving at unit speed.
Let the position be described as a function of time by the function $f(t) = a^{it}$. We want to show that $a=e$
The velocity of the particle is described by $f'(t)$
The condition that velocity is perpendicular to position gives $f'(t) = ikf(t)$ with $k \in \mathbb R$
Since there is no distortion, we must also have $|f'(t)|=1$ so that $k = \pm 1$
So the position function is the solution to $$f'(t) = \pm i f(t)$$
Choosing the positive sign we get $a = e$, the negative sign gives $a = \frac 1e$