The key fact is that:
a rotation of an angle $2\theta$ in space, around an axis passing
through the origin, is represented by a quaternion
$e^{\mathbf{u}\theta}$, where $\mathbf{u}$ is the imaginary quaternion
that correspond to the unit vector oriented along the axis of
rotation. So we have the correspondence:
$$
\vec{w}=R_{\mathbf{u},\theta} \; \vec{v} \quad \longleftrightarrow
\quad \mathbf{w}=
e^{\mathbf{u}\theta/2}\mathbf{v}e^{-\mathbf{u}\theta/2} $$
See my answer to :Quaternions vs Axis angle.
For the exponentiaton we have that if $ \mathbf{v} \in \mathbb{H}_P$ is an imaginary quaternion, putting $\theta=|\mathbf{v}|$ we have:
$$
e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta}
$$
See:Exponential Function of Quaternion - Derivation
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.
Best Answer
The key point you need to get your two orthogonal 2-planes is that the real numbers also form a distinguished line in $\Bbb{H}$.
Suppose first that we have a quaternion $\bf{q}$ which is of the form $\cos \theta + i \sin \theta$. Then multiplication by $\bf{q}$ (either left or right) clearly acts on the span of $\{1,i\}$ by rotating by $\theta$; it's not too hard to check that it also acts on the span of $\{j,k\}$ by rotating by $\theta$ (if you orient $\{j,k\}$ correctly; the correct choice of orientation will depend on whether you're multiplying on the left or on the right). So in this case your two orthogonal 2-planes are the span of $\{1,i\}$ and the span of $\{j,k\}$.
It's much the same for general unit quaternions. In general we have ${\bf q}=\cos \theta + {\bf r }\sin \theta$ where $\bf r$ is a purely imaginary unit quaternion. Then $\bf{q}$ will act on the span of $\{1,{\bf r}\}$ via rotation by $\theta$, and also on its orthogonal complement (but the direction of rotation on the orthogonal complement depends on whether you're left- or right-multiplying).