Verify a Matrix Proof Given $A = A^2$ Problem
Given an arbitrary matrix $ A = A^2 $, prove that $ I - 2A = (I - 2A)^{-1}.$
Attempt
$$ I - 2A = (I - 2A)^{-1} $$
$$ (I - 2A)(I - 2A) = (I - 2A)(I - 2A)^{-1} $$
$$ (I - 2A)(I - 2A) = I $$
$$ (I - 2A)^2 = I $$
This is where I'm not sure if the distributive property can be applied to the following equation. Assuming it can be,
$$ (I - 2A)^2 = I $$
$$ I^2 - 4A + 4A^2 = I $$
$$ I^2 - 4A + 4A = I $$
$$ I^2 = I $$
$$ I = I $$
$$ [LHS] = [RHS] $$
Notes
Is it possible to prove this property is satisfied without using the distributive property as I did above? And what would be an actual example of such a matrix A that satisfies this property?
Solution to Proof
After reviewing and understanding multiple solutions, including that of @JMoravitz, I've come to the following solution.
Consider $ (I - 2A)^{2} $. We must first show that $ (I - 2A)^{2} = I$.
$$ (I - 2A)^2 = (I - 2A)(I - 2A) $$
$$ (I - 2A)^2 = I^2 - 4A + 4A^2 $$
$$ (I - 2A)^2 = I^2 - 4A + 4A $$
$$ (I - 2A)^2 = I^2 $$
$$ (I - 2A)^2 = I $$
Therefore, $ I - 2A $ is the inverse of $ I - 2A $, by the definition of matrix inversion.
An example of such a matrix that satisfies this property is the $ I_{2 \times 2} $ matrix, which is an idempotent matrix.
\begin{bmatrix} 
1 & 0 \\
0 & 1 
\end{bmatrix}
 A: You are told that $A=A^2$ and from this we must prove that $(I-2A)=(I-2A)^{-1}$.
Reworded, letting $P$ be the proposition that $A=A^2$ and letting $Q$ be the proposition that $(I-2A)=(I-2A)^{-1}$ we need to prove that $P\implies Q$
In your attempt, you tried to show that $(Q\wedge P)\implies\textbf{True}$.  This is not the same thing as proving that $P\implies Q$.  You started with what you want to show.  This is not allowed.
Consider for a counterexample, the claim that if $p$ is a prime number then it is equal to $2$.  That is, $p\text{ is prime}\implies p=2$.  You should know that this is a false implication, for example $3$ being a prime number different than $2$.  In your attempt, it is as though you began your proof by saying "*Let $p=2$ (and let $p$ be a prime number).  Then $2=2$, therefore all prime numbers are equal to $2$.*"
Instead, you should begin with your hypotheses and work towards your desired outcome.  (Admittedly it is sometimes useful to work backwards by starting with the desired outcome and performing manipulations, however this is dangerous.  Especially when some of the steps are not biconditional, for example taking square roots or dividing by unknowns which could be zero).

The start of a correct proof.  Suppose that $A=A^2$.  Consider $(I-2A)^2$.  We wish to prove that $(I-2A)^2=I$ which would then imply that $(I-2A)=(I-2A)^{-1}$.
$(I-2A)^2 = (I-2A)(I-2A)=\cdots$

$=I-4A+4A^2=I-4A+4A=I$

To answer your question about whether or not the distributive property works for matrices, it does!  You have for matrices of appropriate dimension $(A+B)(C+D)=AC+AD+BC+BD$.  Be careful however and remember that most matrices will not commute with eachother, so for example $(A+B)(A+B)=A^2+AB+BA+B^2$ is potentially different than $A^2+2AB+B^2$.  Identity matrices do always commute with everything.
A: Yes, you can apply the distributive property with matrices, but be careful with multiplication (you must preserve the order), because it is not commutative, so for example:
$$(A+B)(C+D)=AC+AD+BC+BD$$
An example of a matrix satisfying $A=A^2$ is $ \begin{bmatrix} 2&-2&-4 \\ -1&3&4 \\1&-2&-3 \end{bmatrix}$.
See https://en.wikipedia.org/wiki/Idempotent_matrix
A: Such a linear transformation, $A^2=A$, is called a projection.
For example, an orthogonal projection:
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$.
You seem to have all the elements of a proof that $I-2A=(I-2A)^{-1}$...  
