Quaternions with the restriction that you can only use a single rotation plane work just fine still. They would take the form
$$q = \cos \frac{\theta}{2} + v \sin \frac{\theta}{2}$$
for some unit quaternion $v$.
When we restrict ourselves to a single rotation plane, that means we consider rotating vectors (pure imaginary quaternions) that are perpendicular to the rotation axis (described by $v$).
It is not difficult to show that, for any vector $a \perp v$,
$$\exp(v \theta/2) a \exp(-v \theta/2) = \exp(v\theta) a = a \exp(-v \theta)$$
This is why complex rotations don't usually use the double-sided form that rotations in 3d use: it's simply unnecessary.
From a programming standpoint, the double-sided form requires more floating point operations; this is strictly unnecessary in 2d.
Of course, only the double-sided form of rotation generalizes beyond 3d.
With all this in mind, I think you can consider using quaternions with some terms zeroed out, but notice that when translating quaternions to 2d, the rotation axis is perpendicular to the vectors being rotated. This means you can't zero out particular components regardless of whether they're vectors being rotated or quaternions being used to rotate them.
Furthermore, I think the matter of ensuring some terms remain identically zero could be an issue. You might be better off converting to complex and then converting back when you're done.
Let's work out your proposed mapping:
\begin{align}
e^{j\theta}e^{k\phi}
&= (\cos\theta + j\sin\theta)(\cos\phi + k\sin\phi) \\
&= \cos\theta\cos\phi + (j\sin\theta)(k\sin\phi)
+ (j\sin\theta)\cos\phi + (\cos\theta)(k\sin\phi) \\
&= \cos\theta\cos\phi + i\sin\theta\sin\phi
+ j\sin\theta\cos\phi + k\cos\theta\sin\phi \\
\end{align}
Now, since we know $q = e^{i\alpha/2}e^{k\beta/2}e^{j\gamma/2}$
is a unit quaternion,
it follows that $e^{j\theta}e^{k\phi}$ also is a unit quaternion
(just set $\alpha = 0, \beta = 2\phi, \gamma = 2\theta)$;
but since it has a non-zero real part (except when $\cos\theta=0$
or $\cos\phi=0$),
the imaginary part $ix + jy + kz$ generally will not correspond to a
unit vector in $\mathbb R^3.$
In fact, an even more obvious problem is that if $\theta = 0,$ as $\phi$ varies freely we get a set of vectors whose imaginary parts are all in one direction, while if $\phi = 0,$ as $\theta$ varies we get a set of vectors whose imaginary parts are all in a different direction.
In spherical coordinates we can have only one coordinate such that setting that coordinate to zero forces the vector to be in a particular direction.
But let's consider the interpretation of a quaternion as a rotation.
It is well-known
that a rotation by $\theta$ radians around the axis through the origin
in the direction of the unit vector $\hat{\mathbf u}$ can be represented
by a unit quaternion $q = q_0 + \mathbf q,$
where $q_0 = \cos(\theta/2)$ is real and
$\mathbf q = (\sin(\theta/2))\hat{\mathbf u}$ is the imaginary part of the quaternion interpreted as a vector.
The rotated image $v'$ of a three-dimensional vector $v$ (interpreted as the imaginary part of a quaternion) is computed by $v' = qvq^*,$
where $q^* = q_0 - \mathbf q$ is the conjugate of $q.$
A rotation by $\alpha$ radians around the $x$-axis (or respectively around the $y$- or $z$-axis) is therefore represented by the quaternion
$\cos\frac\alpha2 + i\sin\frac\alpha2 = e^{i\alpha/2}$
(or respectively by $\cos\frac\alpha2 + j\sin\frac\alpha2 = e^{j\alpha/2}$
or $\cos\frac\alpha2 + k\sin\frac\alpha2 = e^{k\alpha/2}$).
Moreover, the conjugate of a quaternion $q$ for the rotation $R$ is a quaternion $q^*$ for the inverse rotation $R^{-1},$ since $q^*q = qq^* = 1,$ which is a rotation by a zero angle (or if you would prefer a different explanation, because reversing the sign of the imaginary part, which is conjugation, is equivalent to reversing the sign of the angle $\theta$ in
$q = \cos(\theta/2) + (\sin(\theta/2))\hat{\mathbf u}$).
Due to the way rotation by means of a quaternion is defined,
the first rotation in a sequence is the rightmost factor in the product of rotation quaternions.
So if we want to rotate the vector $k$ (the unit vector pointing upward along the $z$ axis) by an angle $\phi$ around the $y$-axis and then by an angle $\theta$ around the $z$-axis, the quaternion for this rotation is
$p_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2},$ its conjugate is
$p_{\theta,\phi}^* = e^{-j\phi/2} e^{-k\theta/2},$
and the purely imaginary unit quaternion equivalent to the vector with spherical coordinates $(1, \theta, \phi)$ is
$$
q_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2} k e^{-j\phi/2} e^{-k\theta/2}.
$$
It turns out the answer looks something like your initial guess, with some important distinctions: we use $i\theta/2$ and $i\phi/2$ as exponents instead of $i\theta$ and $i\phi,$ and in order to get the coordinates of the actual desired vector we must use these exponents to rotate the unit vector $k.$
For completeness, here's another way to derive the same answer.
I decided to describe this method later because it requires much more calculation.
Any unit quaternion $ix + jy + kz,$ where $i,$ $j,$ and $k$ are the three imaginary units of the quaternion and $(x,y,z) \in \mathbb R^3,$ represents a $180$-degree rotation around the unit vector $(x,y,z).$
And if
\begin{align}
x &= \sin\phi \cos\theta, \\
y &= \sin\phi \sin\theta, \\
z &= \cos\phi,
\end{align}
then $(1, \theta, \phi)$ are spherical coordinates of the axis of rotation,
and hence the quaternion corresponds almost uniquely to those coordinates.
There is a complication, namely, the $180$ degree rotation around the axis is the same for the axis given by $(x,y,z)$ as for $(-x,-y,-z),$
and likewise the quaternions $q$ and $-q$ represent the same rotation;
hence there is some ambiguity about whether a particular quaternion corresponds to particular spherical coordinates $(1, \theta, \phi)$ or to the coordinates $(1, \theta + \pi, \pi - \phi),$
that is, the antipodes of $(1, \theta, \phi).$
Setting aside that ambiguity for now, the question is just how to construct such a rotation.
One way is to rotate the point at spherical coordinates $(1, \theta, \phi)$
to the positive $z$-axis, rotate by $\pi$ radians around the $z$-axis,
then rotate the positive $z$-axis back to the spherical coordinates
$(1, \theta, \phi).$
The rotations to do this are a rotation by $-\theta$ radians around the $z$-axis, which brings the desired rotation axis into the $x,z$ plane,
followed by a rotation by $-\phi$ radians around the $y$-axis to bring the axis onto the $z$-axis, followed by a rotation by $\pi$ radians around the $z$-axis, then $\phi$ radians around the $y$-axis and finally
$\theta$ radians round the $z$-axis to bring the rotation axis back to its original location.
This rotation described is represented by the quaternion
$$
q_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2} e^{k\pi/2} e^{-j\phi/2} e^{-k\theta/2}.
$$
Working this out in detail, first note that
$e^{k\pi/2} = \cos\frac\pi2 + k\sin\frac\pi2 = k$; then
\begin{align}
e^{j\phi/2} e^{k\pi/2} e^{-j\phi/2}
&= \left(\cos\frac\phi2 + j\sin\frac\phi2\right) k
\left(\cos\frac\phi2 - j\sin\frac\phi2\right) \\
&= k \cos^2\frac\phi2 + jk \sin\frac\phi2\cos\frac\phi2
- kj \cos\frac\phi2\sin\frac\phi2 - jkj \sin^2\frac\phi2 \\
&= k \left(\cos^2\frac\phi2 - \sin^2\frac\phi2\right)
+ 2i \sin\frac\phi2\cos\frac\phi2 \\
&= k \cos\phi + i \sin\phi \\
\end{align}
and so
\begin{align}
q_{\theta,\phi} &= e^{k\theta/2} (k \cos\phi + i \sin\phi) e^{-k\theta/2} \\
&= \left(\cos\frac\theta2 + k\sin\frac\theta2\right)
(k \cos\phi + i \sin\phi)
\left(\cos\frac\theta2 - k\sin\frac\theta2\right) \\
&= \left(k \cos^2\frac\theta2 + k^2\sin\frac\theta2\cos\frac\theta2
- k^2\sin\frac\theta2\cos\frac\theta2
- k^3 \sin^2\frac\theta2\right)\cos\phi \\
&\qquad + \left(i \cos^2\frac\theta2 + ki \sin\frac\theta2\cos\frac\theta2
- ik \cos\frac\theta2\sin\frac\theta2
- kik \sin^2\frac\theta2\right)\sin\phi \\
&= k \left(\cos^2\frac\theta2 + \sin^2\frac\theta2\right)\cos\phi
+ i \left(\cos^2\frac\theta2 - \sin^2\frac\theta2\right)\sin\phi
+ 2j \sin\frac\theta2\cos\frac\theta2\sin\phi \\
&= i \cos\theta\sin\phi + j \sin\theta\sin\phi + k \cos\phi,\\
\end{align}
which is exactly what we want for the unit vector with
spherical coordinates $(1, \theta, \phi).$
Best Answer
I am posting this to get more information about the exact meaning of both $dq$ as well as the product in $\dfrac1q?dq$.
If this is to be a surface integral over the unit sphere, here's how it could go.
The surface of the unit sphere is, indeed, parametrized as $$ q(\theta,\phi)=\cos\theta\cos\phi \mathbf{i}+\cos\theta\sin\phi\mathbf{j}+\sin\theta\mathbf{k} $$ with $\phi\in[0,2\pi)$ growing from West to East and $\theta\in[-\pi/2,\pi/2]$ growing from South to North. I am following your parametrization. Usually the spherical coordinate $\theta\in[0,\pi]$ grows from North to South and measures the latitude difference from the North Pole as opposed to the difference from the Equator which is what your $\theta$ measures.
Anyway, we need the surface element $d\mathbf{S}$ on the unit sphere. It is gotten as the cross product of the derivatives $$ \frac{\partial q}{\partial\theta}=-\sin\theta\cos\phi\mathbf{i}-\sin\theta\sin\phi\mathbf{j}+\cos\theta\mathbf{k} $$ pointing North along a meridian, and $$ \frac{\partial q}{\partial\phi}=-\cos\theta\sin\phi\mathbf{i}+\cos\theta\cos\phi\mathbf{j} $$ pointing East along a latitude. More often than not we orient the surface of the sphres to have an outward pointing normal, so I calculate the cross product in the following order $$ \begin{aligned} d\mathbf{S}&=\frac{\partial q}{\partial\phi}\times \frac{\partial q}{\partial\theta}\,d\phi\,d\theta\\ &=\left(\cos^2\theta\cos\phi\mathbf{i}+\cos^2\theta\sin\phi\mathbf{j}+\cos\theta\sin\theta\mathbf{k}\right)\,d\phi\,d\theta\\ &=\cos\theta\, q\,d\phi\,d\theta. \end{aligned} $$ Therefore (sorry about being worried about non-commutativity of the product of quaternions – doesn't play a role here):
Of course, if we switch the orientation of the surface, then the signs in the two first bullets are swapped. Go figure?