$\mathbf {The \ Problem \ is}:$ We know that the set $S = \{ $$ \begin{bmatrix} x & x \\ x & x \\ \end{bmatrix} $$ | x \in \mathbb R-\{0\} \}$, forms a group under matrix multiplication, with the element $$ \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{bmatrix} $$ as identity and $$ \begin{bmatrix} 1/{4x} & 1/{4x} \\ 1/{4x} & 1/{4x} \\ \end{bmatrix} $$ as inverse for each element . Point to note that each element of $S$ is singular .
Is there any underlying relationship between the standard inverse of a matrix and the group theoretic inverse of the element $$ \begin{bmatrix} x & x \\ x & x \\ \end{bmatrix} $$ ???
$\mathbf {My \ approach} :$ I found the eigenvalues $0 \ and \ 2x$ and the corresponding eigenvectors $(1, -1)$ and $(1,1)$ for each matrix in $S$ and hence we can find an orthogonal eigenvector basis for $\mathbb R^2$ and more interestingly, each element of S can be decomposed uniquely into two invertible matrices ; $$ \begin{bmatrix} x & x \\ x & x \\ \end{bmatrix} = \begin{bmatrix} x & 0 \\ 0 & x \\ \end{bmatrix} + \begin{bmatrix} 0 & x \\ x & 0 \\ \end{bmatrix} $$
Does this fact imply anything ???
Consider the function $f\colon \mathbb R\setminus 0\to S$ given by $$ f(x)=\begin{bmatrix}x/2&x/2\\x/2&x/2\end{bmatrix}. $$ Then $f$ is seen to be a group isomorphism from the multiplicative group of non-zero real numbers to $S$.
In other words, the group is not really using any of the 2-dimensional matrix structure: it is just a copy of our usual group of multiplicative non-zero reals. As you pointed out, none of the matrices in $S$ have a matrix inverse. As such, there is no direct relation between matrix inverses and group inversion in $S$.
By the way, a change of basis might make the situation clearer: this is effectively the same thing as considering the group of matrices $$ T=\left\{\begin{bmatrix}x&0\\0&0\end{bmatrix}\colon x\in \mathbb R\setminus 0\right\} $$ again under matrix multiplication. None of these matrices are invertible, yet we still have a group inverse given by $$ \begin{bmatrix}x^{-1}&0\\0&0\end{bmatrix}. $$