I don't know non-commutative ring theory that well yet, but I ended up working this problem out anyway partially for my own edification. Let us suppose that I, J are both left ideals in R. Observe,

$(I \cap J) (I + J) = (I \cap J) I + (I \cap J) J \subseteq IJ + JI$

If I + J = (1), then we get

$I \cap J \subseteq IJ + JI$

Now it can be shown that

$I J \subseteq comm(I,J) \cap J$

where comm(I,J) denotes the set of all elements in I which commute with elements in J;

$comm(I, J) = ${$ x \in I : \forall y \in J; x y \in I $}

Therefore,

$I \cap J \subseteq IJ + JI \subseteq comm(I,J) \cap J + comm(J,I) \cap I$

If it is true that all of I commutes with J and vice-versa, then $comm(I,J) = I$, $comm(J,I) = J$ and we get:

$I \cap J = I J$

Now let us suppose that these bounds are not tight, or in other words there is some element $x \in I \setminus comm(I,J)$. This implies that there exists some $y \in J$ such that $x y \not \in I$, and so $x J \not \subseteq I \cap J$.

Therefore if $I + J = (1)$, then $I \cap J = I J + J I$ if and only if $comm(I,J) = I$ and $comm(J,I) = J$.

(Of course there is probably an easier way to say this, so maybe it is also good to get an expert to weigh in on the issue.)