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
When does $\text{lcm}(I,J)$ equal $IJ$?
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.
Best Answer
These rings are called simple rings. As Torsten says in the comments, a nice class of examples which are not division rings is the matrix algebras $M_n(k)$. More generally, by the Artin-Wedderburn theorem the artinian simple rings are precisely the rings of the form $M_n(D)$ where $D$ is a division algebra. If the center of $D$ is $k$ then these are central simple algebras over $k$ and are classified by the Brauer group of $k$.
To prove that $M_n(D)$ is simple it's cleaner to prove a more general result:
Proof. Let $X \in M_n(R)$ be any element. The ideal generated by $X$ consists of linear combinations of elements of the form
$$e_{ij} X e_{kl}$$
where $1 \le i, j, k, l \le n$; this matrix has only a single nonzero component, namely the $il$ entry, whose value is $X_{jk}$. So by picking $i, j, k, l$ appropriately we see that we can arrange for any particular component of $X$ to end up in any other component; in other words, the ideal generated by $X$ is $M_n(I)$ where $I$ is the ideal of $R$ generated by the components of $X$. The desired result follows upon taking sums of ideals, which gives more generally that the ideal generated by any collection of matrices is $M_n(I)$ where $I$ is the ideal of $R$ generated by their components. $\Box$
I said above that the artinian simple rings are the rings of the form $M_n(D)$, so let's close with a non-artinian example. Maybe the most famous non-artinian simple rings are the Weyl algebras, of which the one-variable version can be written
$$k[x, \partial]/(\partial x - x \partial = 1)$$
where $k$ is a field of characteristic zero; this can be thought of as the algebra of differential operators on $k[x]$, with $\partial$ acting by differentiation by $x$.
Proof. Let $f = \sum f_{ij} x^i \partial^j$ be an element of the Weyl algebra; we will show directly that if $f$ is nonzero then the ideal it generates is the entire Weyl algebra. First, observe that the monomials $x^i \partial^j$ form a basis of the Weyl algebra; there are various ways to prove this, it is a PBW-type result. So $f = 0$ iff $f_{ij} = 0$ for all $i, j$.
The defining relation of the Weyl algebra can be written $[\partial, x] = 1$, where $[a, b] = ab - ba$ is the commutator bracket. Now, the commutator bracket $[\partial, -]$ is always a derivation, so it follows that
$$[\partial, x^i] = \partial x^i - x^i \partial = ix^{i-1}$$
while $[\partial, \partial] = 0$ and hence $[\partial, \partial^j] = 0$. This gives
$$[\partial, x^i \partial^j] = ix^{i-1} \partial^j$$
and hence
$$[\partial, f] = \sum f_{ij} ix^{i-1} \partial^j.$$
That is, computing the commutator by $\partial$ has the effect of differentiating the polynomial part of $f$ (and note that $[\partial, f] = \partial f - f \partial$ lies in the ideal generated by $f$). So we can repeatedly apply $[\partial, -]$ to $f$ until all of its polynomial parts vanish except the ones of highest degree; hence we can assume WLOG that $f$ in fact has the form
$$f = \sum f_j \partial^j$$
where at least one $f_j$ is zero (this is where we need both the assumption that the $x^i \partial^j$ form a basis and that $k$ has characteristic zero). At this point we can now instead apply the derivation $[-, x]$, which satisfies $[\partial, x] = 1$ and hence by induction
$$[\partial^j, x] = j\partial^{j-1}$$
which gives
$$[f, x] = \sum f_j j \partial^{j-1}.$$
So we can again repeatedly "differentiate" until $f$ is a nonzero constant, which clearly generates the entire Weyl algebra. $\Box$