I finally found a source (see here) which gives the definition: $$\sum_{j \in J} I_j := \left\{ \sum_{j\in J} x_j: x_j \in I_j\ (\forall j)\ \text{and only finitely many }x_j\text{ are nonzero} \right\} .$$
Related to Arnaud D.'s comment, we apparently have the relationship that: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle $$ where $\langle \rangle$ denotes the ideal generated by a set, the intersection of all ideals containing the set.
Thus the set of all proper ideals $I \subseteq R$ of a ring turns out to not only be a partial order under set inclusion, but actually a lattice, meaning that all supremums (least upper bounds) and infimums (greatest lower bounds) exist/are defined.
The infimum of a set of ideals is the arbitrary intersection of the ideals, while the supremum of ideals is the sum of ideals, since it is the smallest ideal containing all of the ideals. Thus the sum and intersection of ideals are indeed somewhat related operations.
Claim: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle $$ Proof: Let $ x \in \sum_{j\in J} I_j$. Then $x=x_1 + \dots + x_n$, with $x_1 \in I_{j_1}, \dots, x_n \in I_{j_n}$, $j_1, \dots, j_n \in J$. Then since $x_1, \dots, x_n \in \bigcup_{i=1}^n I_{j_i}$, we have that $x_1 + \dots + x_n \in L$ for any ideal $L$ such that $L \supseteq \bigcup_{i=1}^n I_{j_i}$. But of course because $\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j$, so: $$\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j \subseteq \bigcap_{J \supseteq \bigcup_{j \in J} I_j} J = \left\langle \bigcup_{j \in J} I_j\right\rangle, \\ L=\left\langle \bigcup_{j \in J} I_j\right\rangle \supseteq \bigcup_{i=1}^n I_{j_i}$$ so $x_1 + \dots + x_n \in \left\langle \bigcup_{j \in J} I_j\right\rangle$.
$x_1 + \dots + x_n \in \sum_{j\in J} I_j$ was arbitrary, we have shown that $\sum_{j\in J} I_j \subseteq \left\langle \bigcup_{j \in J} I_j\right\rangle$.
Since $\sum_{j \in J} I_j$ is an ideal and $\sum_{j \in J} I_j \supseteq \bigcup_{j\in J} I_j$, $$\sum_{j \in J} I_j \supseteq \bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left(\bigcap_{\{J: J \supseteq \bigcup_{j \in J} I _j \} \setminus \{\sum_{j \in J} I_j \}} J\right) \cap \left( \sum_{j \in J} I_j \right).$$ However, $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J$ is literally the definition of $\left\langle \bigcup_{j\in J} I_j \right\rangle$, i.e. $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left\langle \bigcup_{j \in J} I_j \right\rangle$ trivially, we have shown that $$\sum_{j \in J} I_j \supseteq \left\langle \bigcup_{j \in J} I_j \right\rangle$$ In other words, $\left\langle \bigcup_{j \in J} I_j \right\rangle$ is the smallest ideal containing $\bigcup_{j \in J} I_j$, $\sum_{j\in J}I_j$ is an ideal containing the union, so it must contain the smallest ideal containing the union. $\square$
Best Answer
There is a very logical reason why one should require $I+J=R$ to say that $IJ=I\cap J$.
A Dedekind domain $R$ is a ring such that every non-zero proper (all the ideals in the following discussion are assumed non-zero and proper unless stated otherwise) ideal $I$ of $R$ has a unique factorization into prime ideals $I=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}$. Any PID is a Dedekind domain, as well as any number ring (the integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$) such as $\displaystyle \mathbb{Z}\left[\frac{1+\sqrt{-15}}{2}\right]$.
In these rings, the usual ideal-theoretic operations can be almost completely analogized to element-theoretic operations in a UFD. This is largely because of the fact that in a Dedekind domain $R$, one has that $I\subseteq J$ if and only if $J\mid I$ (i.e. there exists some other ideal $J'$ such that $I=JJ'$). From this one can show that the intersection and sum of ideals has a very familiar form. In particular, the intersection of ideals $I\cap J$ plays the role of least common multiple $\text{lcm}(I,J)$. Namely, if
$$I=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}\qquad J=\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_m^{f_m}$$
(where $e_m$ and $f_m$ are integers, possibly zero) then
$$I\cap J=\mathfrak{p}_1^{\max\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\max\{e_m,f_m\}}=\text{lcm}(I,J)$$
Similarly, the sum $I+J$ plays the role of the greatest common divisor $\text{gcd}(I,J)$ so that, in particular if $I$ and $J$ are as above, then
$$I+J=\mathfrak{p}_1^{\min\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\min\{e_m,f_m\}}=\text{gcd}(I,J)$$
So, now we can rephrase your question as
Well, analogizing the case of UFDs, we know what the answer should be: precisely when $\text{gcd}(I,J)=(1)$ (note that the ideal $R=(1)$ plays the role of identity element with respect to multiplication).
But, the above phrasing tells us that $\text{gcd}(I,J)=I+J$. Thus, $I\cap J=IJ$ should occur precisely when $I+J=R$.
Not only does this give us intuition about why we should need $I+J=R$, but in fact allows us to produce many counterexamples. In particular, the above actually shows that $I\cap J=IJ$ if and only if $I+J=R$. So, taking any two non-coprime ideals $I$ and $J$ of a Dedekind domain will produce an example of ideals such that $IJ\subsetneq I\cap J$.
In particular, if we take the Dedekind domain $\mathbb{Z}$, the way we factor ideals into prime ideals is simple. All the non-zero ideals of $\mathbb{Z}$ are of the form $(a)$ where $a\in\mathbb{N}$ and $a\geqslant 2$. But, we can factor $a$ into the product of primes $a=p_1^{e_1}\cdots p_m^{e_m}$ and then
$$(a)=(p_1)^{e_1}\cdots (p_m)^{e_m}$$
is the factorization of $(a)$ into the product of prime ideals. So, we must produce two non-coprime ideals which, by what we just said, amounts to producing two non-coprime elements of $\mathbb{Z}$. Let's take $a=6$ and $b=15$. Then,
$$(a)\cap (b)=\text{gcd}((2)(3),(3)(5))=(2)(3)(5)=(30)$$
but
$$(a)(b)=((2)(3))((3)(5))=(2)(3)^2(5)=(90)$$
So, there's one example.
Let's produce a slightly more sophisticated example. The ring $\displaystyle R=\mathbb{Z}\left[\frac{1+\sqrt{-15}}{2}\right]$ is a Dedekind domain, but not a PID (it has class number $2$ if that means anything to you). But, just like the above all we have to do is produce three prime ideals $\mathfrak{p}_1$, $\mathfrak{p}_2$, and $\mathfrak{p}_3$ and consider the ideals $I=\mathfrak{p}_1\mathfrak{p}_2$ and $J=\mathfrak{p}_2\mathfrak{p}_3$. So, here are three prime ideals of $R$ (I leave it to you to verify their primality):
$$\mathfrak{p}_1=\left(2,\frac{1+\sqrt{-15}}{2}+1\right),\quad \mathfrak{p}_2=\left(3,\frac{1+\sqrt{-15}}{2}\right),\quad \mathfrak{p}_3=\left(5,\frac{1+\sqrt{-15}}{2}\right)$$
So, then the ideals
$$I=\mathfrak{p}_1\mathfrak{p}_2=\left(6,3+\frac{1+\sqrt{-15}}{2}\right),\quad J=\mathfrak{p}_2\mathfrak{p}_3=\left(\frac{1+\sqrt{-15}}{2}\right)$$
Should give us counterexamples. Indeed
$$IJ=\left(30,3\frac{1+\sqrt{-15}}{2}+15\right)$$
and
$$I\cap J=\left(1+\sqrt{-15}\right)$$
and you can check directly that $IJ\subsetneq IJ$, which is what we knew had to happen.
But, as I said, the fact that $IJ=I\cap J$ if and only if $I+J=R$ for a Dedekind domain $R$ allows one to cook up infinitely many examples.