Answer
$S$ is not a basis for $P_2$.
Work Step by Step
Let $S$ be given by $$
S=\left\{\left[\begin{array}{ll}{1} & {0} \\ {2} & {3}\end{array}\right],\left[\begin{array}{cc}{-2} & {1} \\ {-1} & {0}\end{array}\right],\left[\begin{array}{ll}{3} & {4} \\ {2} & {3}\end{array}\right],\left[\begin{array}{rr}{-3} & {-3} \\ {1} & {3}\end{array}\right]\right\}.
$$
Consider the combination
$$a \left[\begin{array}{ll}{1} & {0} \\ {2} & {3}\end{array}\right]+b\left[\begin{array}{cc}{-2} & {1} \\ {-1} & {0}\end{array}\right]+c\left[\begin{array}{ll}{3} & {4} \\ {2} & {3}\end{array}\right]+d \left[\begin{array}{rr}{-3} & {-3} \\ {1} & {3}\end{array}\right]=0, \quad a,b,cd,\in R.$$
Which yields the following system of equations
\begin{align*}
a-2b+3c-3d&=0\\
b+4c-3d&=0\\
2a-b+2c+d&=0\\
3a+3c+3d&=0.
\end{align*}
The coefficient matrix of the above system is given by
$$ \left[ \begin {array}{cccc} 1&-2&3&-3\\ 0&1&4&-3
\\ 2&-1&2&1\\ 3&0&3&3\end {array}
\right].
$$
One can see that the determinant of the coefficient matrix is zero, hence the system has non zero solutions and hence, $S$ is not linearly independent set of vectors. Then, $S$ is not a basis for $P_2$.
