Hint
1) What is $\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}$?
2) The multiplication of matrices is associative.
3) When you are looking for the identity you want
$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} e & e \\ e & e \end{bmatrix}=\begin{bmatrix} x & x \\ x & x \end{bmatrix}$$
Now, do the multiplication on the left, what do you get?
4) With the $e$ from $3)$ solve
$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}=\begin{bmatrix} e & e \\ e & e \end{bmatrix}$$
for $y$. Again, all you need to do is doing the multiplication...
P.S. In order for this to be a group, you need $x \neq 0$.
P.P.S Since $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=2\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, you can prove that
$$F: \mathbb R \backslash\{0 \} \to G$$
$$F(x) =\frac{x}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
is a bijection and it preserves multiplications. Since $\mathbb R \backslash\{0 \}$ is a group it follows that G must also be a group and $F$ is an isomorphism... But this is probably beyond what you covered so far...
Let $A=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}$
Then $AB = \begin{pmatrix} 2 & 2 \\ 1 & 0\end{pmatrix}$.
Clearly, $ab_{12}\neq ab_{21}$. So the set is not closed under multiplication
Best Answer
Take $A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$, $B=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$.
Then $AB \neq BA$. (Note that $\det (AB) \neq 0$.)
Also, $(AB)^T \neq AB$.