Can I do the same things on both sides when I have a not equal sign? I am trying to prove this statement:

If $A$ is not symmetric then $A^{-1}$ is not symmetric 

And the following is my proof
If $A≠A^T$,
\begin{align*}
& A^{-1}A=I≠A^{-1} A^T \\
& I^T \neq \left(A^{-1} A^T \right)^T \\
& I \neq A \left( A^{-1} \right)^T \\
& A^{-1} \neq \left( A^{-1} \right)^T \\
& A^{-1} \neq \left( A^{-1} \right)^T. \\
\end{align*}
I am doing several operations on both sides which usually is justified when there is a equal sign between RHS and LHS. I am not sure if this is valid when there is NOT equal sign. 

As @Daniel Fischer said, this happens to be okay the way you're using it, but it doesn't work in general. See the following counterexample:
$$ \begin{gather*} -1 \neq 1 \\
(-1)^2 \neq (1)^2 \\
1 \neq 1 \\
\text{uh-oh}
 \end{gather*} $$

In particular, applying an operation $f$ to both sides of an inequality will always preserve the inequality if and only if $f^{-1}$ exists. A proof follows.
First we'll show that $f^{-1}$ existing guarantees that the inequality will be held. Take $f$ such that $f^{-1}$ exists and also take $a$ and $b$ such that $ a \neq b $. We now want to show that $f(a) \neq f(b)$. For contradiction, let's assume that $f(a) = f(b)$. Since $f^{-1}$ exists, as we assumed, we may apply it to both sides, giving $f^{-1}(f(a)) = f^{-1}(f(b))$, which is just to say that $a = b$. However, this is a contradiction with $a \neq b$. Thus, assuming that $f(a)=f(b)$ produced a contradiction, so we conclude that $f(a) \neq f(b)$.
Note also that $f^{-1}$ not existing guarantees that inequality won't always be preserved. This is because if $f^{-1}$ doesn't exists, then there are two values $a$ and $b$ such that $a \neq b$ but $f(a) = f(b)$. These two values are an immediate example of when the inequality doesn't hold: though they are unequal, $f$ makes them equal.
