It is not that associativity is required for groups... That is quite backwards: the truth is actually that groups are associative.
Your question seems to come from the idea that people decided how to define groups and then began to study them and find them interesting. In reality, it happened the other way around: people had studied groups way before actually someone gave a definition. When a definition was agreed upon, people looked at the groups they had at hand and saw that they happened to be associative (and that that was a useful piece of information about them when working with them) so that got included in the definition.
If I may say so, it is this which is important to understand. The way we teach abstract algebra nowdays somewhat obscures this fact, but this is how essentially everything comes to be.
To show something is a group you need:
$1.$ Well defined binary operation on a nonempty set.
$2.$ The group operation to be associative.
$3.$ There to be an identity under the operation.
$4$. The set to be closed under the operation.
$5.$ Closure under inverses.
(This might be overkill but early on it is good to justify them all). Alright, here we go...
$1.$ The operation is binary (if you really want you can check well defined). $\mathbb{R}$ is clearly nonempty so our set should be nonempty.
$2$. Is the operation $*$ associative? You need $(a*b)*c=a*(b*c)$.
This operation is associative. Notice that $$ \begin{align} (a*b)*c&=(a+b+ab)*c\\&=(a+b+ab)+c+(a+b+ab)c\\&=a+b+ab+c+ac+ab+abc\\ &= a+(b+c+bc)+a(b+c+bc)\\&=a*(b+c+bc)\\&=a*(b*c)\end{align}$$
$3.$ Is there an identity? That is, is there an $e$ so that $a*e=a$ for all $a$?
Let $b=0$, then $a*0=a+0+a\cdot 0=a+0+0=a$ for any $a \in \mathbb{R}$.
$4.$ Check that $a*b \in \mathbb{R}$ and not $-1$
$a*b=a+b+ab$, but if $a,b \in \mathbb{R}$, then $a+b+ab \in \mathbb{R}$. Done! Now if $a,b \neq -1$, can $a*b=a+b+ab=-1$? Solving for $a$ would yield $a=-1$ and similarly for $b$, but this is impossible as $a,b \neq -1$, so $a*b=a+b+ab\neq -1$ so long as $a,b \neq -1$. Therefore, the set is closed under $*$.
$5.$ That is, if $a \in \mathbb{R}$, we need to find an element $i$ so that $a*i=e$, where $e$ is the identity of the group we found before.
The identity is $0$. We want $i$ such that $a*i=0$. Well, $a*i=a+i+ai=0$. Solving for $i$ gives, $i=-\frac{a}{a+1}$, so if $a\neq -1$, we can always find an inverse. Notice if we define $i$ as before then $$a*i=a+\frac{-a}{a+1}+a\frac{-a}{a+1}=\frac{(a^2+a)-a-a^2}{a+1}=0$$.
But then we've shown this set under $*$ is a group! Moreover, we can show this group is commutative:
$$
\begin{align}
a*b&=a+b+ab \\
&=b+a+ba \\
&=b*a
\end{align}
$$
because $a,b \in \mathbb{R}$ and addition and multiplication are commutative in $\mathbb{R}$.
Best Answer
Indeed we have to show that the set is closed under
additionin each case. For example, consider $2/3$ and $1/3$. Both of them satisfy $|.|<1$ but $|2/3+1/3|$ is not strictly smaller than $1$. Socdoesn't make a groupid either. The similar story can be seen fordfor example. In fact $|3/2+(-4/2)|<1$ while $|3/2|\ge1,~~|-4/2| \ge1$.