Closures are trivial. Let's list what we have to prove:
i) Associativity;
ii) Existence of identity;
iii) Existence of inverses.
For $(\Bbb R, \ast)$, you should see what $(a \ast b) \ast c$ and $a \ast (b \ast c)$ are. What you have done does not makes it clear that they're equal. We have: $$\begin{align} (a \ast b) \ast c &= (a + b + ab)\ast c = a+b+ab+c + ac+bc+abc \\ a \ast (b \ast c) &= a \ast (b + c + bc) = a+b+c+bc+ab+ac+abc\end{align}$$
hence they're equal. So we've checked i). For the identity, we want $e \in G$ such that $a \ast e = e \ast a = e$ for all $a \in G$. So we want: $$a+e+ae= a \implies e + ae = 0 \implies e(1+a) = 0$$
Now, $e = 0$ will work for every $a$, but if $a = -1$, we have a problem here, because, calling $a^{-1}$ a candidate for the inverse of $-1$, we get: $-1 \ast a^{-1} = -1 + a^{-1} + (-1)a^{-1} = 0$, and so $-1 = 0$ by cancelling $a^{-1}$, a contradiction. Hence $-1$ doesn't have an inverse, and $(\Bbb R, \ast)$ is not a group. Property iii) fails.
Now, let's see the other one.
Consider $(\Bbb Z^2, \ast)$. If we want to see that $$\left((a,b)\ast(c,d)\right) \ast (q,r) = (a,b) \ast \left((c,d) \ast (q,r)\right)$$ then we should see each side separately, and then compare. We have: $$\begin{align} \left((a,b)\ast(c,d)\right)\ast (q,r) &= (ad+bc, bd)\ast(q,r) = (adr+bcr + bdq,bdr) \\ (a,b) \ast \left((c,d) \ast (q,r)\right) &= (a,b) \ast (cr + dq, dr) = (adr + bcr + bdq, bdr)\end{align}$$ hence property i) holds. For ii), we want $(e_1, e_2)$ such that $(a,b) \ast (e_1, e_2) = (e_1, e_2) \ast (a,b)$, that is: $(ae_2 + be_1, be_2) = (a,b)$, which gives $e_2 = 1$. And that gives $a + be_1 = a$, so $e_1 = 0$. To check that $(0,1)$ really is the identity, see that $(0,1)\ast(a,b) = (0b + a, 1b) = (a,b)$. So, property ii) also holds. Now, let $(\bar{a}, \bar{b})$ be a candidate for the inverse of $(a,b)$. We want: $(a,b) \ast (\bar{a}, \bar{b}) = (0,1)$. That is: $$(a\bar{b} + b\bar{a}, b\bar{b}) = (0,1)$$
This already gives us problems, since if you choose $b = 0$, the inverse will not exist. Hence $(\Bbb Z^2, \ast)$ is not a group, because property iii) fails.
To prove it is the unique solution, you need to prove two things:
Your solution is a solution, i.e. you do indeed need to show that your solution is associative, that $e$ is indeed the identity element, and that every element has an inverse.
There are no other solutions.
Now, to show that there are no other solutions, you would go through an argument like:
$e$ being the identity element immediately forces 5 of the 9 values in the table:
\begin{array}{|c|c|c|c|c|c|}
\hline
*& e & a & b \\ \hline
e& e&a &b\\ \hline
a& a& &\\ \hline
b& b & &\\ \hline
\end{array}
$a$ has an inverse element, i.e. there is some element $a^{-1}$ such that $a * a^{-1} = e$, so $a^{-1} \not = e$, since $a * e = a$ and $a \not = e$.
Now, if $a^{-1} = a$ is inverse of $a$, then $aa=e$, so $(b * a) * a = b * (a * a) = b * e = b$, and so $b * a = b$, since if $b * a = e$ then $(b * a) * a = e * a = a \not = b$, and if $b * a = a$ then $(b * a) * a = a * a = e \not = b$. But since $b$ has to have an inverse $b^{-1}$ such that $b * b^{-1} = e$, and since $b * e = b \not = e$, and $b * a = b \not = e$, that means that $b^{-1} = b$. But then $(a * b) * b = b * b = e$,while $a * (b * b) = a * e= a$, and so $*$ is not associative. Hence, we have a contradiction if $a^{-1}= a$, and hence the only way things can work is if $a^{-1}= b$, i.e. $a * b = e$
\begin{array}{|c|c|c|c|c|c|}
\hline
*& e & a & b \\ \hline
e& e&a &b\\ \hline
a& a& &e\\ \hline
b& b & &\\ \hline
\end{array}
But this means that $a*a = b$, for we have $(a*a)*b = a * (a * b) = a * e = a$, and if $a*a = e$ then $(a * a) * b = e * b = b \not = a$, and if $a * a = a$ then $(a * a) * b = a * b = e \not = a$.
\begin{array}{|c|c|c|c|c|c|}
\hline
*& e & a & b \\ \hline
e& e&a &b\\ \hline
a& a& b&e\\ \hline
b& b & &\\ \hline
\end{array}
Also, since $a * (b * a) = (a * b) * a = e * a = a$, we get that $b * a = e$, for if $b * a = a$ then $a * (b * a) = a * a = e \not = a$, and if $b * a = b$, then $a * (b * a) = a * b = e \not = a$.
\begin{array}{|c|c|c|c|c|c|}
\hline
*& e & a & b \\ \hline
e& e&a &b\\ \hline
a& a& b&e\\ \hline
b& b & e&\\ \hline
\end{array}
And finally, since $a * (b * b) = (a * b) * b = e * b = b$, it must be that $b * b = a$, for if $b * b = e$, then $a * (b * b) = a * e = a \not = b$, and if $b * b = b$, then $a * (b * b) = a * b = e \not = b$
\begin{array}{|c|c|c|c|c|c|}
\hline
*& e & a & b \\ \hline
e& e&a &b\\ \hline
a& a& b&e\\ \hline
b& b & e&a\\ \hline
\end{array}
OK, so what we have now established is that there cannot be more than 1 solution.
But we have not yet established that this actually is a solution!
That is, if you look back at the proof, you see that we rules out all kinds of values, but we never established that the values that were forced actually did satisfy the requirements!
So, yes, you need to show that $e$ works like an identity element (easy), that ever element has an inverse (easy), and that this $*$ is associative (not hard, but tedious)
Best Answer
In number $4$ we have $da=d$ and $de=d$. Now in a group, $d^{-1}$ exists, and it follows $a=e$ (from $da=d=de$), a contradiction. Hence it is not a group.