A noncommutative counterexample to the following property: If $I,J$ are comaximal ideals, then $IJ=I\cap J$.

It suffices to find ideals $I,J$ such that $I+J=R$ and $IJ \neq JI$. In fact, then $IJ = I \cap J$ and $JI = J \cap I$ cannot hold both, so that either $(I,J)$ or $(J,I)$ is a counterexample, or both, as below.

Consider $R=\mathbb{Z}\langle x,y \rangle / (2x+2y=1)$, $I=(x)$, $J=(y)$. Then $I+J=R$, and one can prove that $I \cap J = (xy,yx)$, which strictly contains $IJ=(xy)$. (In order to simplify calculations, you may work with $R/(xy)$.)

Here is a positive result:

If $I,J$ are two-sided ideals of a ring $R$ with $I+J=R$, then $IJ+JI = I \cap J$.

Proof: The inclusion $IJ +JI \subseteq I \cap J$ follows from $IJ \subseteq I$, $IJ \subseteq J$, $JI \subseteq I$, $JI \subseteq J$. Conversely, $I \cap J = (I \cap J)R = (I \cap J)(I+J) = (I \cap J)I + (I \cap J)J \subseteq JI+IJ$. $\checkmark$