$AB \subseteq A \cap B$ is clear. I have seen reverse inclusion proven thus,
Let $x \in A\cap B$. Since $A+B=R$, there exist $a \in A$, $b \in B$, such that $a+b=1$. Then $x= axa + axb + bxa + bxb$. Therefore, $A \cap B \subseteq AB$.
My problem: I cannot see why $bxa \in AB$.
Edit: If this does not hold for noncommutative rings, give a counterexample
I will modify your reverse inclusion statement. Let us suppose we have $a+b = 1$ for $a \in A$ and $b \in b$. Then for $x \in A \cap B$ we have $xa+xb = x$. Note that $xa \in AB$ because $x \in A \cap B$ (see note below) and similarly $xb \in AB$. Thus $x$ is the sum of two elements in $AB$. Thus, $x \in AB$, and so we have reverse inclusion.
As mentioned by Potato above a Commutative Ring is needed here (I use it implicitly when I say $xa \in AB$, because $xa$ is really in $BA$, but by being in a commutative ring, $BA = AB$).
As a counterexample, take the ring of $2 \times 2$ upper triangular matrices over $\mathbb{R}$. Take $A = \begin{pmatrix} 1 & 0\\ 0 & 0 \\ \end{pmatrix}$, $B = \begin{pmatrix} 0 & 0\\ 0 & 1 \\ \end{pmatrix}$, $I = (A), J = (B)$ (as left ideals). Now all matrices of $I$ are of the form $ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}$ and the matrices of $J$ are of the form $ \begin{pmatrix} 0 & c \\ 0 & d \end{pmatrix}, \forall a,b,c ,d \in \mathbb{R}$. Now $A+B = Id$, so $I+J = R$. Yet, $I \cap J = 0$ and if $C = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \in J$, $AC = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in IJ$.