The key observation is that swapping the first and third column yields
$$
\frac{1}{\sqrt2}\begin{bmatrix}
0 & 1 & 1 & 0 \\
0 & -1 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1
\end{bmatrix} \to \frac{1}{\sqrt2}\begin{bmatrix}
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & -1
\end{bmatrix} = I\otimes H\tag1
$$
where $H=\frac{1}{\sqrt2}\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}$ is the Hadamard gate. The coefficients $\frac{1}{\sqrt2}$ are needed to ensure unitarity. The column swap corresponds to the permutation matrix
$$
P = \begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}\tag2
$$
which maps $|00\rangle$ to $|10\rangle$ and $|10\rangle$ to $|00\rangle$. In other words, it flips the first qubit when the second qubit is in the $|0\rangle$ state. This is the usual controlled-NOT gate with two modifications: the first qubit is the target and the second is the control and the gate acts non-trivially if the control qubit is in the $|0\rangle$ state rather than the usual $|1\rangle$. Thus,
$$
P = (I\otimes X)\circ\text{CNOT}_{2,1}\circ(I\otimes X).\tag3
$$
Putting it all together, we get
$$
\frac{1}{\sqrt2}\begin{bmatrix}
0 & 1 & 1 & 0 \\
0 & -1 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1
\end{bmatrix} = (I\otimes H) \circ (I\otimes X)\circ\text{CNOT}_{2,1}\circ(I\otimes X)\tag4
$$
where the last expression is written in the linear algebraic convention, i.e. the corresponding gates are executed from right to left.
