From Wikipedia:
Given a ring R and a two-sided ideal I in R, we may define an equivalence relation ~ on R as follows:
a ~ b if and only if a − b is in I.Using the ideal properties, it is not difficult to check that ~ is a congruence relation. In case a ~ b, we say that a and b are congruent modulo I. The equivalence class of the element a in R is given by
[a] = a + I := { a + r : r in I }.This equivalence class is also sometimes written as a mod I and called the "residue class of a modulo I".
The set of all such equivalence classes is denoted by R/I; it becomes a ring, the factor ring or quotient ring of R modulo I, if one defines
(a + I) + (b + I) = (a + b) + I; (a + I)(b + I) = (a b) + I.
Suppose that we no longer restrict $I$ to be an ideal, but just any set that is a subset of $R$ and is not an ideal of $R$. Then would it be impossible to satisfy (a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = (a b) + I all the time if $I$ is not an ideal?
No. You can't take just any subset $I$ of $R.$ To define quotient ring, first it required to be a group. So the subset $I$ has to be a subgroup of $(R, +).$ (Actually we require normal subgroup to define quotient group, but in this case $(R,+)$ is abelain. So any subgroup is normal.) Now take $I$ to be a subgroup of $(R, +)$ and consider the quotient group $R/I.$ We want to show that it is a ring where the multiplication is defined by $(x+I)(y+I)=xy+I.$ First we need to make sure that it's a well defined operation. Suppose $a+I=b+I.$ Then for every $c+I \in R/I,$ we need $(c+I)(a+I)=(c+I)(a+I),$ i.e. $ca+I=cb+I$ and so $ca-cb\in I.$ Now $a+I=b+I \Rightarrow a-b \in I.$ But why should $c(a-b) \in I, \forall c \in R$? There we need the property of ideal: $x\in I, y \in R \Rightarrow yx \in I.$
To emphasis the last point, let's consider the following example: $R$ be the ring of all functions from $\mathbb R$ to $\mathbb R.$ Let $C$ be the subset of $R$ containing only the constant functions. Then $C$ is a subgroup of $R,$ but not an ideal of $R.$ Here we have $x+C=x+2+C.$ Multiplying both sides by $x+C,$ we would like to have $x^2+C=x^2+2x+C.$ But this is not true because $x^2 - (x^2 +2x) \notin C.$