New answer
Let $\mathbb H$ be the $\mathbb R$-algebra generated by the symbols $i_0$ and $j_0$ subject to the relations $ i_0^2=-1=j_0^2,\quad i_0\ j_0=-j_0\ i_0. $
Let $X$ be an indeterminate and define the $\mathbb R[X]$-algebra $\mathbb H[X]$ by $ \mathbb H[X]:=\mathbb R[X]\underset{\mathbb R}{\otimes}\mathbb H. $ We must find a maximal left ideal $\mathfrak a$ of $\mathbb H[X]$ which is not an ideal.
Put $ \mathbb C:=\frac{\mathbb R[X]}{(X^2+1)} $ and denote by $x\in\mathbb C$ the canonical image of $X$.
Note that the left ideal $\mathfrak b\subset\mathbb H[X]$ generated by $X^2+1$ is an ideal. In particular $ A:=\frac{\mathbb H[X]}{\mathfrak b} $ is a $\mathbb C$-algebra.
If $\pi:\mathbb H[X]\to A$ is the canonical projection, and if $\mathfrak c$ is a maximal left ideal of $\mathbb H[X]$ which is not an ideal, then $\mathfrak a:=\pi^{-1}(\mathfrak c)$ fits the bill.
Let $M$ be the $\mathbb C$-algebra of two by two matrices with entries in $\mathbb C$, and let $i,j\in A$ be the canonical images of $1\otimes i_0,1\otimes j_0\in\mathbb H[X]$.
One easily checks that there is a unique $\mathbb C$-algebra morphism $\phi:A\to M$ satisfying $ \phi(i)=\begin{pmatrix}x&0\\0&-x\end{pmatrix},\quad \phi(j)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, $ and that $\phi$ is bijective.
It is also easy the verify that the left ideal of $M$ generated by $ e:=\begin{pmatrix}1&0\\0&0\end{pmatrix}=\phi\left(\frac{1-ix}{2}\right) $ is maximal, and that it is not an ideal.
This implies that we can put $ \mathfrak c:=A\ \frac{1-ix}{2}\quad, $ or, in other words, that $ \pi^{-1}\left(A\ \frac{1-ix}{2}\right) $ is a maximal left ideal of $\mathbb H[X]$ which is not an ideal.
Old answer
Let $\mathfrak a$ be the ideal of $\mathbb H[X]$ generated by $X^2+1$, let $A$ be the quotient $\mathbb H[X]/\mathfrak a$.
It suffices to find a maximal left ideal of $A$ which is not an ideal.
Write $x\in A$ for the canonical image of $X$. Let us identify $\mathbb R[x]$ to $\mathbb C$. Then $A$ is isomorphic, as a $\mathbb C$-algebra, to $M_2(\mathbb C)$.
(See the very beginning of the text of Gaëtan Chenevier named Lecture 6 on this html page. Here is a direct link to the pdf file.)
As a result, $A$ contains an idempotent $e$ with $1\neq e\neq0$, and $Ae$ is the sought-for maximal left ideal.
EDIT. One can put $ e:=\frac{1+ix}{2}\quad. $ (I use the traditional notation $i,j,k$ instead of the $I,J,K$ in the question.)