Is set $S$ with these operations a ring? 
Consider a set $S=\{a,b\}$ with addition and multiplication defined by:
  $$\begin{align*}
a+a&=a& a+b&=b& b+a&=b& b+b&=a\\
a\times a&=a& a\times b&=a& b\times a&=a& b\times b&=b
\end{align*}$$
Is $S$ a ring?

If it is a ring since addition and multiplication that would be done are all in the sets itself would that be the correct answer? 
 A: The most straightforward to verify that it is a ring is to interpret "a" as "0", and "b" as "1". Then S has the ring structure of the integers, mod 2.
A: Assuming multiplication is associative, but might not be commutative or have an identity, there are only two rings with two elements. These are $\mathbb{F}_2 = \mathbb{Z} / 2\mathbb{Z}$, and $\{0,1\}$ where addition is mod $2$ and multiplication always returns $0$.
So, you just have to check if the addition and multiplication tables match either of these two rings. As others have observed, you should quickly see it matches $\mathbb{F}_2$.
A: 
If it is a ring since addition and multiplication that would be done are all in the sets itself would that be the correct answer? 

No, the set being closed under addition ("being all in the set itself") is not sufficient for it being a ring. It is,of course, necessary (so if it were not the case, it couldn't be a ring).
To check that you have a ring,you have to check that all the ring axioms are fulfilled. Of course, as the other answers mentioned, you can shortcut the tests by mapping to structures you already know to have the desired property.
So if you already know that $\mathbb Z/2\mathbb Z$ is a ring, it is sufficient to show that you can identify $a$ and $b$ with $0$ and $1$ and obtain $\mathbb Z/2\mathbb Z$. If you only know that $\mathbb Z/2\mathbb Z$ is a group under addition, you can use that identification, too, but still need to prove the ring properties involving multiplication.
