Answer
$S$ does not span $M_{2,2}$.
Work Step by Step
The set
$S=\left\{\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right],\left[\begin{array}{ll}{1} & {1} \\ {0} & {0}\end{array}\right]\right\}$
is not a basis for $M_{2,2}$ because $S$ does not span $M_{2,2}$. For example, consider the matrix $\left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]$ as a linear combination of the elements of $S$ as follows
\begin{align*}
\left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]&=a\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]+c\left[\begin{array}{ll}{1} & {1} \\ {0} & {0}\end{array}\right]\\
&=\left[\begin{array}{ll}{a+c} & {b+c} \\ {b} & {a}\end{array}\right].
\end{align*}
We have the following system of equations
\begin{align*}
a+c&=1\\
b+c&=0\\
a&=0\\
b&=0.
\end{align*}
It is clear that the above system is in-consistent. Hence, $S$ does not span $M_{2,2}$.
