Answer
$S$ is a basis for $M_{2,2}$.
Work Step by Step
Let $S$ be given by $$
S=\left\{\left[\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right],\left[\begin{array}{cc}{-1} & {0} \\ {1} & {1}\end{array}\right],\left[\begin{array}{cc}{2} & {1} \\ {1} & {0}\end{array}\right],\left[\begin{array}{cc}{1} & {1} \\ {0} & {1}\end{array}\right]\right\}.
$$
Consider the combination
$$a \left[\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{cc}{-1} & {0} \\ {1} & {1}\end{array}\right]+c\left[\begin{array}{cc}{2} & {1} \\ {1} & {0}\end{array}\right]+d \left[\begin{array}{cc}{1} & {1} \\ {0} & {1}\end{array}\right]=0, \quad a,b,cd,\in R.$$
Which yields the following system of equations
\begin{align*}
a-b+2c+d&=0\\
c+d&=0\\
b+c&=0\\
a+b+d&=0.
\end{align*}
The coefficient matrix of the above system is given by
$$ \left[ \begin {array}{cccc} 1&-1&2&1\\ 0&0&1&1
\\ 0&1&1&0\\ 1&1&0&1\end {array}
\right]
.
$$
One can see that the determinant of the coefficient matrix is non zero, hence the system has only the trivial solution and hence, $S$ is linearly independent set of vectors. Since, $M_{2,2}2$ has dimension $4$, then by Theorem 4.12 $S$ is a basis for $M_{2,2}$.
