I have thot about this more, with the help of the comments above, which I will cite in my answer.
The answer is no, moving the vertical bar is not a valid matrix operation, because:
- It is not a matrix operation at all (Henning Makholm)
- If it were to be defined as a matrix operation, it would be rather ambiguous (Ted)
However, the underlying matrix operations that were behind the idea are valid. The idea is just as
$$\left[\begin{array}{ccc|c}
1 & 0 & -1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]$$
implies
$$\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}\right]
\left[\begin{array}{c}
x \\ y \\ z
\end{array}\right] =
\left[\begin{array}{c}
1 \\ 0 \\ 0
\end{array}\right]$$
So
$$\left[\begin{array}{cc|cc}
1 & 0 & 1 & 1 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0
\end{array}\right]$$
would imply
$$\left[\begin{array}{ccc}
1 & 0 \\
0 & 1 \\
0 & 0
\end{array}\right]
\left[\begin{array}{c}
x \\ y
\end{array}\right] =
z \left[\begin{array}{c}
1 \\ -1 \\ 0
\end{array}\right]
+ \left[\begin{array}{c}
1 \\ 0 \\ 0
\end{array}\right]$$
which is legal, and solves the problem of switching in and out of the matrix notation and should provide a good intermediate step, as desired, even if it is not quite as succinct. It also makes the next steps much more clear: do the matrix multiplication on the LHS, add
$$ \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} $$
to both sides of the equation, and simplify.