Is a matrix equivalent with its row reduced one? If I manage to reduce a matrix A to the identity one, does that mean that I can actually use  it on any given equation instead?
$$AB = 0 \implies IB = 0\;\;?$$
 A: A row-reduced matrix is row-equivalent to the original matrix, but not equal to it.
In your example, if you are able to row reduce a matrix $A$ to obtain the identity matrix, it is invertible, and non-zero. So in you example, you CAN conclude that $$AB = {\bf 0} \iff A^{-1}AB = A^{-1}{\bf 0} \implies B = {\bf 0}$$.
ADDED: What is true is that a system of equations, when represented by an augmented coefficient matrix, and then row-reduced using Gaussian Elimination, the row reduced system of equations is equivalent to the original system of equations. Any solution  to the system of equations represented by the row reduced matrix is a solution to the original system of equations. But a square matrix, in and of itself, does not represent a system of equations. 
Consider an invertible square matrix $A$. $$\begin{bmatrix} 2 & 0 \\ 0&2\end{bmatrix}$$By definition, as an invertible matrix, this can be row-reduced to the identity matrix (obvious in this case). To see that the original and the reduced matrix are not equivalent, note that the determinant of $A$ is $4$, whereas the determinant of $I = 1$.
A: Counterexample:
Note that
$$
A = \pmatrix{
1&1&0\\
0&1&1\\
0&0&1
}
$$
Can be reduced to the identity.  Taking $b = \pmatrix{2\\2\\1}$
We note that the unique solution to 
$Ax = b$ is given by $x = \pmatrix{1\\1\\1}$.  Replacing $A$ with $I$, however, would give us the equality $x = b$, which is not the case.
