Hint: for arbitrary $a \in \mathbb{R} \setminus \{0\}$, we're after a solution to the equation
$$\begin{bmatrix}
a & a \\
a & a \\
\end{bmatrix}
\begin{bmatrix}
b & b \\
b & b \\
\end{bmatrix}
=
\mathrm{id}_G
=
\begin{bmatrix}
1/2 & 1/2 \\
1/2 & 1/2 \\
\end{bmatrix}$$
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...
Best Answer
Let $S$ be a set of $2\times 2$ matrices defined by:
$$S \, = \, \left\{\begin{pmatrix} a & c \\ b & d\end{pmatrix} \; : \; a+b=1, \, c+d=1, \, ad\neq bc\right\}.$$
Then $A\in S$ is of the form
$$A = \begin{pmatrix} a & c\\ 1-a & 1-c \end{pmatrix}$$
where $a, c\in\mathbb{R}$ with $a\neq c$. Now take two matrices in $A$:
$$A = \begin{pmatrix} a & c \\ 1-a & 1-c \end{pmatrix}, \quad B = \begin{pmatrix} a' & c' \\ 1-a' & 1-c' \end{pmatrix}. $$
with $a, c, a', c' \in\mathbb{R}$ with $a\neq c, \, a'\neq c'$. For closure, we require $AB\in S$:
$$AB = \begin{pmatrix} a & c \\ 1-a & 1-a \end{pmatrix}\begin{pmatrix} a' & c' \\ 1-a' & 1-c' \end{pmatrix} \, =\, \begin{pmatrix} aa' +(1-a')c & ac' +(1-c')c \\ (1-a)a' + (1 - c) (1 - a') & (1-a)c' + (1 - c)(1-c') \end{pmatrix} $$
which after simplifying gives
$$AB = \begin{pmatrix} aa' +(1-a')c & ac' +(1-c')c \\ 1-[aa'+(1-a')c] & 1 - [ac' +(1-c')c] \end{pmatrix} $$
which is of the same form (elements in first column add to 1, as do the elements in the second column; elements in first row can't be equal since $a\neq c$) and $AB\in S$. Hence $S$ is closed under matrix multiplication. That should get you started.