Answer
See below
Work Step by Step
$A1$. Suppose $(x_1,y_1),(x_2,y_2) \in R^2$
Then $(x_1,y_1) \oplus (x_2,y_2)=(x_1+x_2,y_1+y_2) \in R^2$
Hence, $V$ is closed under addtion.
$A2$. Suppose $(x_1,y_1) \in R^2, c \in R$
Then $c(x_1,x_2)=(cx_1,cx_2)$
Hence, $V$ is closed under scalar multiplication.
$A3$. Suppose $(x_1,y_1),(x_2,y_2) \in V$
Then $(x_1,y_1) \oplus (x_2,y_2)=(x_1+x_2,y_1+y_2)=(x_2+x_1,y_2+y_1)=(x_2,y_2) \oplus (x_1,y_1)$
$A4$. Suppose $(x_1,y_1),(x_2,y_2),(x_3,y_3) \in V$
then $(x_1,y_1) \oplus (x_2,y_2) \oplus (x_3,y_3)=((x_1+x_2).(y_1+y_2))\oplus (x_3+y_3)=((x_1+x_2)+x_3, ((y_1+y_2)+y_3)=(x_1+(x_2+x_3),y_1+(y_2+y_3))=(x_1,y_1)+(x_2+x_3,y_2+y_3)=(x_1,y_1) \oplus ((x_2,y_2) +(x_3,y_3))$
$A5$. Suppose $x \in R^+$
then $(x_1,y_1)+(0,0)=(x_1+0,y_1+0)=(x_1,y_1)$
Hence, $0$ is zero vector
$A6$. Suppose $(x_1,y_1) \in V$
then $(x_1,y_1) \oplus (-x_1,-y_1)=(x_1-x_1,y_1-y_1)=(0,0)$
Hence, additive inverse of $x$ is $(-x_1,-y_1)$.
$A7$. Suppose $(x_1,y_1) \in V$
then $1 \oplus (x_1,y_1)=(1.x_1,1.y_1)=(x_1,y_1)$
$A8$. Suppose $(x_1,y_1) \in V$ and $r,s \in R^+$
then $(rx) \oplus (x_1,y_1)=(rs(x_1,y_1))=r(sx_1,y_1)=(rsx_1,y_1)=rs \odot (s \odot (x_1,y_1)$
$A9$. Suppose $(x_1,y_1),(x_2,y_2) \in V$ and $r \in R$
then $r \odot (x_1,y_1 \oplus x_2,y_2)=r \odot (x_1+x_2, y_1+y_2)=(r(x_1+x_2),y_1+y_2)=(rx_1+rx_2,y_1+y_2)=(rx_1,y_1)+(rx_2,y_2)=r \odot (x_1,y_1) \oplus r \odot (x_2,y_2)$
$A10$. Suppose $(x_1,y_1) \in V$ and $r,s \in R$
then $(r+s)\odot (x_1,y_1)=((r+s)x_1,y_1)=(rx_1+y_1)+(sx_1,y_1)=r \odot (x_1,y_1) \oplus s\odot (x_1,y_1)$
