Proving $A$ is invertible if $A + A^2 = I$.

From $A+A^2=I$ you do indeed get that $AA'=I$ for some $A'$, namely $A'=I+A$. And since the distribution laws hold "on both sides", you also get $A'A=I$ for the same $A'$, which proves that $A$ is invertible.

With $A(I+A)=I$ you’re almost there: this shows that $A$ has a right inverse, $I+A$. But a square matrix has a right inverse if and only if it’s actually invertible, so $A$ is invertible.

If you don’t already know that theorem, you can prove it in a variety of ways. For instance, $$I=I^T=\big(A(I+A)\big)^T=(I+A)^TA^T\;,$$ so $A^T$ has a left inverse. Suppose that $A^Tx=0$. Then $$x=Ix=(I+A)^TA^Tx=(I+A)^T0=0\;,$$ so the null space of $A^T$ is trivial, containing only the zero vector. Therefore $A^T$ is invertible, and its inverse must be $(I+A)^T$. Thus, $I=A^T(I+A)^T=\big((I+A)A\big)^T$, and therefore $$(I+A)A=I^T=I\;,$$ meaning that $I+A$ is also a left inverse of $A$. Since $I+A$ is both a left and a right inverse of $A$, it’s actually $A^{-1}$, and $A$ is invertible.

Added: As tst points out in the comments, I was working much too hard here. We can factor the $A$ out of $A+A^2$ on the right as well as on the left, so not only do we have $A(I+A)=I$, we also have $(I+A)A=I$, and it’s immediate that $I+A=A^{-1}$.

We have $1=\det{(I)}=\det{(A+A^2)}=\det{(A(I+A))}=\det({A)}\det{(I+A)}$. Thus $\det({A)}\neq 0$ and $A$ is invertible.