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
You really just have to write out the two expressions $(x*y)*z$ and $x*(y*z)$. If you do, you get: \begin{align*}(x*y)*z&=((x*y)^{1/3}+z^{1/3})^{3}\\ &=(\left((x^{1/3}+y^{1/3})^3\right)^{1/3} +z^{1/3})^3\\ &=(x^{1/3}+y^{1/3}+z^{1/3})^3.\end{align*} You can write out the other expression, and see that the same cancellation takes place and that they are equal.
There is a more general fact at play here however: The map $f:\mathbb R\rightarrow \mathbb R$ given by $f(x)=x^{1/3}$ is a bijection. If you already know that addition is commutative and associative, you can show the same of this operation if you note that $$f(x*y)=f(x)+f(y)$$ Since then you can write $$f(x*y)=f(x)+f(y)=f(y)+f(x)=f(y*x)$$ and then, since $f$ is a bijection, that means $x*y=y*x$. Similarly, for associativity, you can just write $$f((x*y)*z)=f(x*y)+f(z)=f(x)+f(y)+f(z)=f(x)+f(y*z)=f(x*(y*z))$$ therefore $(x*y)*z=x*(y*z)$.
This is a nice way to view this sort of question, since it tells you that $*$ basically has the exact same properties as the usual addition on $\mathbb R$. More formally, these operations are isomorphic because they satisfy this relation, and this means they are essentially the same, except that we "relabelled" the points in $\mathbb R$ somehow.