(a) To check injectivity, we compute the kernel of $S$. To this end, suppose that $a+bx+cx^2+dx^3\in\krn{S}$, so
\begin{equation*}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
=\lt{S}{a+bx+cx^2+dx^3}
=
\begin{bmatrix}
-a+4b+c+2d & 4a-b+6c-d\\
a+5b-2c+2d & a+2c+5d
\end{bmatrix}
\end{equation*}
this creates the homogeneous system of four equations in four variables,
\begin{align*}
-a+4b+c+2d&=0\\
4a-b+6c-d&=0\\
a+5b-2c+2d&=0\\
a+2c+5d&=0
\end{align*}
The coefficient matrix of this system row-reduces as,
\begin{align*}
\begin{bmatrix}
-1 & 4 & 1 & 2 \\
4 & -1 & 6 & -1 \\
1 & 5 & -2 & 2 \\
1 & 0 & 2 & 5
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & 0 \\
0 & \leading{1} & 0 & 0 \\
0 & 0 & \leading{1} & 0 \\
0 & 0 & 0 & \leading{1}
\end{bmatrix}
\end{align*}
We recognize the coefficient matrix as being nonsingular, so the only solution to the system is $a=b=c=d=0$, and the kernel of $S$ is trivial, $\krn{S}=\set{0+0x+0x^2+0x^3}$. By Theorem KILT, we see that $S$ is injective.
(b) We can establish that $S$ is surjective by considering the rank and nullity of $S$.
\begin{align*}
\rank{S}
&=\dimension{P_3}-\nullity{S}&&\knowl{./knowls/theorem.RPNDD.knowl}{\text{Theorem RPNDD}}\\
&=4-0\\
&=\dimension{M_{22}}
\end{align*}
So, $\rng{S}$ is a subspace of $M_{22}$ (Theorem RLTS) whose dimension equals that of $M_{22}$. By Theorem EDYES, we gain the set equality $\rng{S}=M_{22}$. Theorem RSLT then implies that $S$ is surjective.
(c) Since $S$ is both injective and surjective, Theorem ILTIS says $S$ is invertible.