proves that $AA^{-1} = I$ for matrix $A$ below
$A_{n _\times n} = \begin{pmatrix} B_{r _\times r} & C_{r _\times s} \\ D_{s _\times r} & E_{s _\times s} \\ \end{pmatrix}$
and $r+s=n$
I found the inverse in such a way that:
$A^{-1} = \begin{pmatrix} (B-CE^{1}D)^{-1} & -B^{-1}C(E-DB^{-1}C)^{-1} \\ -E^{-1}D(B-CE^{-1}D)^{-1} & (E-DB^{-1}C)^{-1} \\ \end{pmatrix}$
but I'm spinning around and I can not complete this proof
I cannot comment, so this becomes an answer:
Inverses of $2\times 2$ block matrices
Matrix Inversion Lemmas
(and many other links found by the search engine) give related computations for the problem, showing the problems with computations in non-commutative rings. The above formula, rewritten to have notations already burned on my chip, holds after we assume that all built inverses exist. This will be tacitly assumed in the following. Let $$ X = \begin{bmatrix} A & B\\\\C&D \end{bmatrix}\ . $$ We consider $$ Y = \begin{bmatrix} S & T \\\\ U & V \end{bmatrix}\ , $$ where \begin{align} S &= (A-BD^{-1}C)^{-1}\ ,\\\\ T &= -A^{-1}B(D-CA^{-1}B)^{-1}=-A^{-1}B\; V\ ,\\\\ U &= -D^{-1}C(A-BD^{-1}C)^{-1}=-D^{-1}C\; S\ ,\\\\ V &= (D-CA^{-1}B)^{-1}\ . \end{align} Then \begin{align} \text{We want } AS+BU &= I\ .\\\\ &\qquad\text{Equivalently, after right multiplication with } S^{-1}=(A-BD^{-1}C)\\\\ A+B(-D^{-1}C ) &= (A-BD^{-1}C)\ .\\\\ &\qquad\text{This is true.} \\\\ % \text{We want } AT+BV &= 0\ .\\\\ &\qquad\text{Equivalently, after right multiplication with } V^{-1}=(D-CA^{-1}B)\\\\ A(-A^{-1}B ) +B&= 0\ .\\\\ &\qquad\text{This is true.} \\\\ % \text{We want } CS+DU &= 0\ .\\\\ &\qquad\text{Equivalently, after right multiplication with } S^{-1}=(A-BD^{-1}C)\\\\ C+D(-D^{-1}C )&= 0\ .\\\\ &\qquad\text{This is true.} \\\\ % \text{We want finally } CT + DV&= I\ .\\\\ &\qquad\text{Equivalently, after right multiplication with } V^{-1}=(D-CA^{-1}B)\\\\ C(-A^{-1}B )+D &= (D-CA^{-1}B)\ .\\\\ &\qquad\text{This is true.} \\\\ \end{align}