By induction on $k$ one can easily reduce the question to $k=2$.

Claim If $\mathscr I, \mathscr J\subseteq \mathscr O_X$ are two ideal sheaves such that $\mathscr I +\mathscr J =\mathscr O_X$, then $\mathscr I\mathscr J=\mathscr I\cap \mathscr J$.

Proof First, observe that the condition implies that

\begin{equation}\tag{1} \mathscr O_X/(\mathscr{I\cap J}) \simeq (\mathscr O_X/\mathscr I) \oplus (\mathscr O_X/\mathscr J). \end{equation}

Second, observe that since $\mathrm{supp} \mathscr I\cap \mathrm{supp}\mathscr J=\emptyset$, we have that \begin{equation}\tag{2} \mathscr{I/(IJ)}\simeq \mathscr I\otimes (\mathscr O_X/\mathscr J) \simeq \mathscr O_X\otimes (\mathscr O_X/\mathscr J) \simeq \mathscr O_X/\mathscr J \end{equation}

Next consider the following commutative diagram with the "obvious" natural maps

\begin{align} 0 &\to & \mathscr I/(\mathscr I\mathscr J) & \to &\mathscr O_X/(\mathscr{IJ}) & \to & \mathscr O_X/\mathscr I & \to & 0 \\ & & \downarrow\qquad\ & & \downarrow\qquad\ & & \downarrow\quad\ & \\ 0 &\to & \mathscr I/(\mathscr I\cap\mathscr J) & \to &\mathscr O_X/(\mathscr{I\cap J}) & \to & \mathscr O_X/\mathscr I & \to & 0 \\ &&& & \downarrow\qquad\ & \\ &&& & \mathscr O_X/\mathscr J. \end{align}

By $(1)$ the induced map $\mathscr{I/(I\cap J)}\to \mathscr O_X/\mathscr J$ is an isomorphism of $\mathscr O_X$-modules and by $(2)$ the induced map $\mathscr{I/(IJ)}\to \mathscr O_X/\mathscr J$ is also an isomorphism of $\mathscr O_X$-modules.

This implies that the natural map $$ \mathscr{I/(IJ)}\to \mathscr I/\mathscr{(I\cap J)} $$ is an isomorphism and hence $\mathscr{IJ=I\cap J}$.

Here is a simple proof for $k=2$. Karl's examples show that the statement is false in general (without more careful assumptions).

Claim If $\mathscr I, \mathscr J\subseteq \mathscr O_X$ are two ideal sheaves such that $\mathscr I +\mathscr J =\mathscr O_X$, then $\mathscr I\mathscr J=\mathscr I\cap \mathscr J$.

Proof $$ \qquad \qquad\qquad \mathscr{(I+J)(I\cap J)}\subseteq \mathscr{I J}\subseteq \mathscr{I\cap J}. \qquad\qquad\qquad \square $$