Answer
See below
Work Step by Step
$A1$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix},B=\begin{bmatrix} e & f \\ g & h
\end{bmatrix} \in M_2(R)$
Then $A \oplus B=A.B=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh
\end{bmatrix}\in M_2(R)$
Hence, $M_2(R)$ is closed under addtion.
$A2$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \in M_2(R)$ and $k \in R$
Then $c.A=cA=\begin{bmatrix} a & b \\ c & d
\end{bmatrix}=A=\begin{bmatrix} ka & kb \\ kc & kd
\end{bmatrix}$
Hence, $M_2(R)$ is closed under scalar multiplication.
$A3$. Suppose $A,B \in M_2(R)$
Then $A \oplus B=A.B=B.A=B \oplus A$
$A4$. Suppose $A,B,C \in M_2(R)$
then $(A \oplus B) \oplus C=(A.B).C=A.(B.C)=A \oplus (B \oplus C)$
$A5$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \in M_2(R)$
then $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \oplus \begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix}=\begin{bmatrix} a & b \\ c & d
\end{bmatrix}.\begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix}=\begin{bmatrix} a & b \\ c & d
\end{bmatrix}$
Hence, $\begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix}$ is zero vector
$A6$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \in M_2(R)$
then $A \oplus (-A)=A.(-A)=\begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix}$
Matrix $-A$ exists only if $rank (A)=2$
Hence, $A6$ is not satisfied.
$A7$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \in M_2(R)$
then $1.A=1.\begin{bmatrix} a & b \\ c & d
\end{bmatrix}=\begin{bmatrix} 1a & 1b \\ 1c & 1d
\end{bmatrix}=\begin{bmatrix} a & b \\ c & d
\end{bmatrix}=A$
$A8$. Suppose $A=\begin{bmatrix} a & b \\ c & d
\end{bmatrix} \in M_2(R)$ and $r,s \in R$
then $(rs)A=(rs)\begin{bmatrix} a & b \\ c & d
\end{bmatrix}=\begin{bmatrix} (rs)a & (rs)b \\ (rs)c & (rs)d
\end{bmatrix}=\begin{bmatrix} r(sa) & r(sb) \\ r(sc) & r(sd)
\end{bmatrix}=r\begin{bmatrix} sa & sb \\ sc & sd
\end{bmatrix}=r(sA)$
$A9$. Suppose $A,B \in M_2(R)$ and $r \in R$
then $r(A \oplus B)=r(A.B)=r.A.B$
and $rA \oplus rB=(rA).(rB)=r^2.A.B$
Hence, $r.A.B \ne r^2.A.B$
$A10$. Suppose $A \in M_2(R)$ and $r,s \in R$
then $(r+s)A=rA+sA$
and $rA \oplus sA=rA.sA$
We can see that $rA + sA \ne rA.sA$
