One matrix has two distinct row echelon forms but one reduced row echelon form

The replacement $R_1\leftarrow R_1+(-1)R_0$ is an elementary row operation of the third kind and takes $A$ to $R$; the scaling $R_1\leftarrow(-1)R_1$ is one of the second kind, $-1$ being nonzero in $\mathbb Q$, and takes $R$ to $S$. So $A$ is row equivalent to each of $R$ and $S$ by [L3]. Each has no zero row, leading entries in columns $0$ and $1$ respectively, and $0$ below the leading entry of its first row, so each is in row echelon form by [L2]. Finally $R\ne S$ because $-1\ne1$ in $\mathbb Q$. The ground field matters here and not elsewhere in the example: over $\mathbb F_2$ one has $-1=1$, so $R$ and $S$ coincide and this matrix exhibits no failure of uniqueness.

From $S$, the replacement $R_0\leftarrow R_0+(-1)R_1$ is again elementary of the third kind and gives $I_2$, so $I_2$ is row equivalent to $A$ by [L3] and step 1.1; its two leading entries equal $1$ and each is alone in its column, so it is in reduced row echelon form by [L2]. Applying the same replacement after the scaling of step 1.1 carries $R$ to $I_2$ as well, in agreement with the uniqueness of [L1].