As a more explicit version of Starfall's answer, in $n\times n$ case, if you denote the standard basis of matrix space by $(E_{i,j})$, you can define a basis simply by $E'_{1,1}=I$, $E'_{i,j}:=I+E_{i,j}$ if one of $i,j$ is not $1$. (You could also take simply $E'_{1,1}:=I+E_{1,1}$, but that would make it harder to show that $(E_{i,j})$ form a basis.)
It is very easy to check that each $E'_{i,j}$ is invertible (just calculate the determinant), the span of $(E'_{i,j})$ contains all $E_{i,j}$ (this is immediate except for $E_{1,1}$, and for $E_{1,1}$ it follows from the observation for $i=j>1$), and since there are $n^2$ of $(E'_{i,j})$, they must form a basis (because the space of matrices is $n^2$-dimensional).
In $2\times 2$ case, the basis is
$$
\left[ \begin{matrix}
1 & 0 \\
0 & 1
\end{matrix} \right], \left[ \begin{matrix}
1 & 1 \\
0 & 1
\end{matrix} \right], \left[ \begin{matrix}
1 & 0 \\
1 & 1
\end{matrix} \right], \left[ \begin{matrix}
1 & 0 \\
0 & 2
\end{matrix} \right]
$$
(Note that more or less by the same argument as the one cited by Starfall, if you just pick $n^2$ random $n\times n$ matrices (for any reasonable notion of "random", but you can think of normally, independendently distributed coefficients, or independently uniformly distributed in a fixed interval), they will all be invertible and they will be linearly independent, and hence form a basis. This is because the set of non-invertible matrices is very small, as is the set of vectors in a proper subspace of a given real vector space.)