For ex,
$$A^{-1}\cdot A=\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]\cdot\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$
and
$$A\cdot A^{-1}=\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]\cdot\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$
so we can make two systems:
\begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \end{cases} \end{equation*}
and
\begin{equation*} \begin{cases} x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}
after line addition we get the system:
\begin{equation*} \begin{cases} 2x_2 + 5x_4 = 1, \\ 2x_1 + 5x_3 = 1, \\ 3 x_1 + 4 x_2 = 1, \\ 3 x_3 + 4 x_4 = 1 \end{cases} \end{equation*}
We solve the system and find the roots: $(x_1 = -2, x_2 = -2, x_3 = 1, x_4 = 1)$. This is not the inverse matrix. What is obvious. The inverse matrix found by the standard algorithm is equal to
$$A^{-1}=\left[\begin{matrix}3 & -2\\-1 & 1\end{matrix}\right]$$
But where are my arguments wrong?
Edited correct roots of the last system are: $x_1 - \frac{10x_4}{3} = \frac{-1}{3}, x_2 + 2.5x_4 = 0.5, x_3 + \frac{4x_4}{3} = \frac{1}{3}$
It is even easier that what you have done. Consider $$A^{-1}\cdot A=\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]\cdot\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$
You get your system: \begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \end{cases} \end{equation*}
Solve it, it is easy to see: $(x_1,x_2,x_3,x_4)=(3,-2,-1,1)$.
You can do the same in the other way. Consider $$A\cdot A^{-1}=\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]\cdot\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$
You get your system: \begin{equation*} \begin{cases} x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}
Solve it, it is easy to see: $(x_1,x_2,x_3,x_4)=(3,-2,-1,1)$.
Hence $$A^{-1}= \left[\begin{matrix}3 & -2\\-1 & 1\end{matrix}\right]$$
Your problem comes when adding these two systems. If you have your new system:
\begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \\ x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}
I think that you have done some bad addition between these equations. Because it should lead the same answer as you have seen above.