Another way of putting it: if $A$ is positive definite, then so is the $2 \times 2$ submatrix consisting of rows and columns $i$ and $j$ (for any $i \ne j$). The determinant of this submatrix is $a_{ii} a_{jj} - a_{ij}^2$, and if the entries are in $\{0,1\}$ the only way to have $a_{ii} a_{jj} - a_{ij}^2 > 0$ is $a_{ii} = a_{jj} = 1$ and $a_{ij} = 0$.
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
Matrix multiplication in general is distributive. If $A$ is $m\times n$ matrix and $B,C$ are $n\times p$ matrices then for all $1\leq i\leq m,1\leq j\leq p$ we have:
$(A(B+C))_{ij}=\sum_{k=1}^n A_{ik}(B+C)_{kj}=\sum_{k=1}^n A_{ik}B_{kj}+\sum_{k=1}^n A_{ik}C_{kj}=(AB)_{ij}+(AC)_{ij}$
So $A(B+C)=AB+AC$. From here you can easily prove what you wanted to prove. So I don't think it has something to do with identity, assuming I understood your question right.