Let $D_+ = \partial_x +x, D_-=-\partial_x+x$. $$D= \begin{pmatrix} 0 & D_- \\ D_+ & 0 \end{pmatrix} $$ which acts on a dense subspace $C_c(\Bbb R) \oplus C_c(\Bbb R)$ of $L^2(\Bbb R) \oplus L^2(\Bbb R)$. Then
$$\frac{1}{\sqrt{(1+D^2)}} D: L^2(\Bbb R)^2 \rightarrow L^2(\Bbb R)^2$$ is a well defined operator.
I do not understand why this statement is true. I know that $D$ is essentially self adjoint.
It means that the operators $(D \pm iI)^{-1}$ are bounded and defined on all of $L^2(\Bbb R)^2$. So that that we may apply functional calculus to the $C^*$ algebra $B(L^2(\Bbb R)^2)$ and perhaps define the square root of their product.
However, the problem is making sense of the product. I am not familiar with the theory of Sobolev spaces.
Since $D$ is essentially self-adjoint, its closure, $\bar{D}$ is self-adjoint. You can define your operator by applying the bounded functional calculus for $\bar{D}$ to the function $x\mapsto\frac{x}{\sqrt{1+x^2}}$.
It is interesting to note that the bounded operator you obtain is the product of the two operators $\bar{D}$ and $(1+\bar{D}^2)^{-\frac{1}{2}}$ (in this order) and an extension of the operator $(1+\bar{D}^2)^{-\frac{1}{2}}\bar{D}$ defined on the domain of $\bar{D}$.