Let G be a multiplicative group of $n\times n$ matrices. If $Int(G)\not = \emptyset$ then $G$ is open on $\mathbb{R}^{n^2}$

You're pretty close: you want to "move" the point $y$ to be near $a$, but you can't just multiply it by $a$; you have to multiply it by the inverse of $b$, too. However, it's easier not to wrap it in a proof by contradiction.

Suppose that $B(b,r) \subseteq G$. Note that "multiplication by $b$" is a homeomorphism (with inverse "multiplication by $b^{-1}$). It follows that $b^{-1}B(b,r)$ is also open, and it is an open set containing $I_n$. And then for any $a \in G$, also $ab^{-1}B(b,r)$ is an open set, now containing $a$. Since every point of $G$ has a neighbourhood in $G$, the set $G$ is open.

You're very close. Here's how I would proceed: let $a\in G$ and $b \in \textrm{int}(G)$ with its associated ball $B(b,r)$. Since the multiplication action of $G$ on itself is transitive, there is a $g\in G$ such that $gb = a$. Now, since translation is a homeomorphism in a topological group, $gB(b,r)$ is an open set containing $a$. Show $gB(b,r)$ lies in $G$ and you're done!

I should note that the solution has nothing to do with the fact that $G$ is a group of matrices. Can you generalize?