Remember from groups and vector spaces the notions of "sub-group" and of "sub-space": you must have a
subset which is also "closed" with respect to the operations. That is, if you take elements from the subset, and you subject them to the operations you have, you should end up with results that are also in your subset.
With groups, that means that if a subset $S$ of a group $G$ is going to be a group, then it must be the case that whenever $s,t\in S$, then $st\in S$ ("closed under multiplication"; if you take two things in $S$ and multiply them, you don't need to go "outside of $S$" to find the answer, it is all already inside $S$; no need to go outside); and if $s\in S$, then you need $s^{-1}\in S$ ("closed under inverses"); and you need $e\in S$ (closed under the operation that gives the identity element).
With vector spaces, if a subset $W$ of a vector space $V$ is going to be a subspace, you need that $\mathbf{0}\in W$ (closed under the operation that gives the identity element); that if $w_1,w_1\in W$, then $w_1+w_2\in W$ (closed under vector addition); and that if $w\in W$ and $\alpha\in F$, then $\alpha w\in W$ (closed under scalar multiplication).
The same is true for rings: if you have a ring $R$, then in order for a subset $S$ to be a subring you need it to be (i) closed under addition; (ii) closed under additive inverses; (iii) contain the identity element of the sum; and (iv) closed under multiplication (if $a,b\in S$, then $ab\in S$). Saying "is a subgroup" is the same as saying (i), (ii), and (iii).
Ideals, however, are a bit more than simply subrings. They play exactly the analogous role to rings as normal subgroups play to groups. You may recall that a normal subgroup $N$ of a group $G$ is a "subgroup-with-something-extra": not only is it "closed" under the three usual operations of the group (identity element, addition, and inverses), it must also be "closed" under a bunch of other operations (conjugation).
Similarly with ideals: not only must they be "closed" under the four usual operations of the ring (additive identity element, addition, additive inverses, and multiplication), it must also be "closed" under a bunch of other operations (that are like the scalar multiplications in the vector space case): left multiplication by any element of $R$, and right multiplication by any element of $R$. Just like "normal subgroup" is "subgroup with something extra", likewise ideals are "subrings with something extra."
The problem with saying "an ideal is a subring that is closed under left and right multiplication" is that saying "closed under left and right multiplication" only says that if $a,b\in I$, then $ab\in I$; you really want to add the coda of "...left and right multiplication by elements of the ring."
Examples abound. Take $R=\mathbb{R}[x]$, the polynomials with coefficients in the real numbers, and consider $\mathbb{Z}[x]$, the polynomials with integer coefficients. The latter is a subggroup of $R$, and if $p(x),q(x)\in\mathbb{Z}[x]$, then the product $p(x)q(x)$ is also in $\mathbb{Z}[x]$. So $\mathbb{Z}[x]$ is a subring of $R$. It is not, however, an ideal of $R$, because $x\in\mathbb{Z}[x]$, and $\pi\in R$, but their product $\pi x\notin\mathbb{Z}[x]$.
Or take $S=\mathbb{R}[x^2]$, the polynomials with real coefficients in which only even powers of $x$ occur. This is a subring of $R$ but not an ideal, because $x^2 \in S$, $x\in R$, but $x(x^2)\notin S$.
On the other hand, $S=\{p(x)\in R\mid p(0)=0\}$, the collection of all polynomials with constant term equal to $0$, is both a subring and an ideal: if $p(x)\in S$ and $q(x)\in R$, then $q(x)p(x)\in S$ since $q(0)p(0)=0$ (and $p(x)q(x)=q(x)p(x)$). That is, $q(x)S\subseteq S$ and $Sq(x)\subseteq S$.