For some background information, given sets $I_1,\dots,I_n \subset A$ of a $C^{*}$ algebra $A$, we define $\prod_{k=1}^n I_k$ to be the closed linear span of all products of the form $\prod_{k=1}^n a_k$ for $a_k \in I_k$.
The claim is that given a $C^{*}$ algebra $A$ and closed ideals $I,J \subset A$, we have the equality $I \cap J = IJ$. The proof is simple enough.
To show $IJ \subset I \cap J$ is obvious. If $a \in IJ$, then $a=bc$ for $b \in I$ and $c \in J$. Since $I,J$ are ideals, $bc \in I \cap J$, so $a \in I \cap J$.
To show $I \cap J \subset IJ$, it suffices to show that $(I \cap J)^{+} \subset IJ$, since $(I \cap J)^{+}$ linearly spans the $C^{*}$ algebra $I \cap J$ (closed ideals of a $C^{*}$ algebra are self adjoint.)
Let $a \in (I \cap J)^{+}$. Then by definition, $a^{\frac{1}{2}} \in I \cap J$. Since $I$ is a $C^{*}$ algebra itself there exists an approximate unit $(u_{\lambda})_{\lambda \in \Lambda} \subset I$. Hence: $$a = \lim_{\lambda} u_{\lambda}a = \lim_{\lambda} u_{\lambda}a^{1/2}a^{1/2} = \lim_{\lambda} (u_{\lambda}a^{1/2})a^{1/2}.$$ Since $u_{\lambda}a^{1/2} \in I$ for all $\lambda \in \Lambda$, and the net $(u_{\lambda}a^{1/2})_{\lambda \in \Lambda}$ converges to $a^{1/2}$, it follows that $a \in IJ$.
My question is, why do we need to introduce an approximate unit into the mix? Why can't we just say since $a^{1/2} \in I \cap J$, and $a=a^{1/2}a^{1/2}$, that $a^{1/2} \in I$ and $a^{1/2} \in J$, so $a^{1/2}a^{1/2}=a \in IJ$?
From your approach, I see that you already know that closed ideals are $C^*$-algebras.
In that case, your proof is essentially correct and does not require the use of an approximate unit. Note however that when you say $a \in IJ$, then $a=bc$ for $b\in I$ and $c \in J$, you did make a tiny mistake. In general, such an element $a$ is a limit of sums of such products.
In reply to a comment above:
$a \in (I\cap J)^+$ means that $a$ is a positive element in the $C^*$-algebra $I \cap J$ , so it is certainly positive in both the ideals $I$ and $J$. Since $C^*$-algebras are closed under taking square roots of positive elements, we get $a^{1/2}\in I \cap J$.