So in the complex picture, rotations and dilations of the plane are acheived just by multiplying with a complex number.
I see here you are transferring that idea, but as you said, you get stuck if your three axes are $1,i,j$, because $ij=k$, which is a fourth axis perpendicular to the first two. If you're willing to imagine a four-dimensional space, then you have a similar picture to the complex numbers: multiplication by $i$ would permute the set $\{\pm 1, \pm i,\pm j,\pm k\}$ just as multiplying by $i$ permutes $\{\pm 1, \pm i\}$.
There is a more useful way to view quaternions as rotations in 3-space, though. Imagine $i,j,k$ acting as orthogonal unit vectors, as in physics. Linear combinations of these model every point in that 3-space as a "pure" quaternion $v$ with no real part. Now given another nonzero quaternion $q$, conjugating $v$ to become $qvq^{-1}$ produces a tranformation of the 3 space! When $q$ is a unit-length quaternion, this transformation is actually a rotation.
Try it out with $q=i$ and see what happens to the $i,j,k$ axes! If you feel like skipping $j$ and $k$, you can try $\sqrt{2}/2 +i\sqrt{2}/2$ (whose inverse is, naturally, $\sqrt{2}/2 -i\sqrt{2}/2$.)
This is not useful in the complex numbers because $aba^{-1}=b$ by commutativity, so no change occurs. However, the noncommutativity of the quaternions allows interesting stuff to happen :)
One thing to keep in mind about this conjugation action modeling rotations is that many rotations are produced by a pair of unit-length quaternions (and not just one unit-length quaternion). All that means is that there is a slight redundancy in the way unit-length quaternions model rotations.
EDIT:
I just wanted to add a little more trying to justify why "just one more" doesn't seem to work. In analogy with the complex numbers, you want to get $1,i,j$ such that the produce of any two lands in $\{\pm1, \pm i \pm j\}$. As you noted, we have to figure out where $ij$ goes. But if $ij=\pm i$, that implies that $-j=\pm -1$, and then $j$ is not orthogonal to 1. Similarly, $ij\neq \pm j$. The only thing left is if $ij=\pm 1$, but then $-j=\pm i$, and $j$ is not perpendicular to $i$.
To briefly address your specific question about "duality" first, there is no "duality" (not sure what is precisely meant by that term here) between complex exponentiation and complex multiplication, unless the arguments of both are real. This is because complex multiplication is commutative; complex exponentiation is not - neither is real exponentiation $(e^{ix}$ is not the same as $ix^{e}$). Take the second line:
$$c^{a+bi}\cdot z \leftrightarrow (a+bi)\cdot z
$$
$$
\text{stretching and rotating} \leftrightarrow \text{stretching and rotating}
$$
Yes, $(a+bi)\cdot z$ corresponds to a transformation consisting of some stretching and some rotation, but the stretching is due to both multiplying by $a$ and multiplying by $bi$.
Now, let's break down intuitively what different operations on the complex plane are (I know you said you already went over some of this in your question, but I think it will lead into the explanation for complex exponentiation nicely).
Complex addition is the same as vector addition: we add the components. Think of this intuitively by imagining each complex number as an arrow: adding two complex numbers is like sticking one of their arrows on the end of the other. Another way: think of adding a complex number not as a static operation, but as a transformation. Adding the number $(a+bi)$ is the same as shifting the origin of the complex plane onto the point $(-a-bi)$. Take a second to imagine why that's true. Thus $(a+bi)$ is a function with respect to addition, which maps every point in the complex plane to another point in the complex plane, (a+bi) away.
Complex multiplication corresponds to both a stretch and a rotation (usually). $$(a+bi)\cdot(c+di)=(ac-bd)+(ad+bc)$$
A better way to think about this is in terms of Euler's formula: represent your two complex numbers as polar coordinates, and multiplication becomes much clearer: $$r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}$$
So we can imagine complex multiplication of two numbers is taking the angles they make with the x axis, adding those two angles to get the angle of your new number, then multiplying the magnitudes of the two original numbers to get the magnitude of your new number. Think of complex multiplication by $(a+bi)$ as two, very dynamic transformations composted together: first stretching the entire complex plane by a factor of $\sqrt{a^2+b^2}$, then rotating the entire complex plane by a factor of $\tan^{-1}(\frac{b}{a})$. Thus $(a+bi)$ can also be thought of as a function with respect to multiplication: it maps every point in the complex plane to another point in the complex plane by combination of a rotation and a stretch.
What kind of function is "complex exponentiation"? We define it as follows: $a^{b+c\cdot i}$=$a^{b}\cdot a^{c\cdot i}$ where $a^{c \cdot i} = e^{c\cdot i \log a}$, etc. based on Euler's formula. $e^{c\cdot i \log a}$ is $e$ raised to a complex number and is thus a rotation. Note that we can imagine complex exponentiation again as a sort of dynamic transformation, squishing and mapping the complex plane to a new location.
Let's break this down into two cases. The first is the only one your question asks about.
1) The base of the exponent is real. As can be seen from above, every "complex exponentiation" is a transformation of the complex plane consisting of two transformations: first, a stretch by some factor, and then a rotation. This is very similar to complex multiplication, which begs the question, when do these two things behave in the same way? You called it "duality," I'm not going to call it that because that word means something specific in linear algebra, I'll just call this similarity "similarity."
Multiplying $x$ by $(a+bi)\rightarrow$ Stretch by $\sqrt{a^2+b^2}$, rotate by $\tan^{-1}(\frac{b}{a})$ radians.
Raising $x$ to the $(a+bi)$ power $\rightarrow$ Stretch by $x^a$, rotate by $b\cdot i \log x$ radians.
Note that the two are very dissimilar - exponentiation is dependent on the base whereas multiplication is not. This is a result of the effect that multiplication by a complex number is something called a linear transformation (https://en.wikipedia.org/wiki/Linear_map), raising something to a complex power is most certainly not.
2) The base of the exponent is complex. This gets a little more complicated, because raising a complex number to a real power corresponds in part to a rotation, so separating which parts are the rotation and which parts are the stretching is a little annoying and won't give much insight here. Complex exponentiation of complex numbers is really funky, giving rise to all sorts of weird fractal shapes when we consider which complex numbers get really large when we raise them to a complex power, and which ones don't - this is related to how the Mandelbrot set is formed (https://en.wikipedia.org/wiki/Mandelbrot_set). The point is that, again, the magnitude of the stretch and rotation is dependent on x, meaning that this is not a linear transformation.
If you want to gain an intuition for how complex exponentiation of complex numbers works, I recommend you play around with some functions with this grapher: http://davidbau.com/conformal/#z
So, to return to what I mentioned at the beginning, the reason why you're not seeing a "duality" in case 3 is that there's no "duality" in case 2 to begin with - yes, both complex exponentiation and complex multiplication correspond to both a rotation and a stretch, but the rotation and stretch for each behave in very different ways. Complex multiplication is a linear transform, complex exponentiation is not.
I also enjoy brilliant.org's courses; if you're interested, I would recommend you check out their course on linear algebra next (https://brilliant.org/courses/linear-algebra/). This is the first answer I've actually posted. I would love feedback from anyone if they have it.
Best Answer
There is a really important aspect of complex numbers that depends on the complex plane having exactly this shape: complex multiplication.
Complex numbers can not only be characterized in cartesian coordinates by a real part and an imaginary part, but also in polar coordinates by a length and an angle.
You know that for any $z \in \mathbb{C}$ there exist $x, y \in \mathbb{R}$ such that $z = x+i\cdot y$, right? $x$ is the real part and $y$ is the imaginary part? Well, there also exist $r, \varphi \in \mathbb{R}$, $r \geq 0$ such that $z = r\cdot(\cos\varphi + i\sin\varphi)$. Here, $r$ is called the length or absolute value of $z$ and $\varphi$ is called the angle or argument, measured counterclockwise from the positive real axis.
We can use cartesian coordinates to add complex numbers: $$(x_1+i\cdot y_1) + (x_2+i\cdot y_2) = (x_1+x_2) + i\cdot(y_1+y_2)$$
We can use cartesian coordinates to multiply complex numbers:
$$(x_1+i\cdot y_1)\cdot (x_2+i\cdot y_2) = (x_1x_2-y_1y_2) + i\cdot(x_1y_2+y_1x_2)$$
However, we can also use polar coordinates to multiply complex numbers:
$$(r_1(\cos\varphi_1 + i\sin\varphi_1))\cdot(r_2(\cos\varphi_2 + i\sin\varphi_2)) = (r_1\cdot r_2)(\cos(\varphi_1+\varphi_2) + i\sin(\varphi_1+\varphi_2))$$
So to multiply two complex numbers in polar coordinates, you multiply their lengths and add their angles. I personally think this is incredibly helpful for visualization, and this also shows why the imaginary axis needs to be at a right angle to the real axis: since the angle of $-1$ is $180^\circ$, the angle of $i$ needs to be $90^\circ$ or $270^\circ$.