The operation you mentioned
$$\Delta \hat{a} = \hat{A}-\langle \hat{A}\rangle \hat{I}$$
is just shifting of the annihilation operator. Typically people even drop the identity and write
$$\hat{a}_{new}=\hat{a}_{old}-\alpha_0,$$
where $\alpha_0$ is a complex number. In your question the shift is such that $\langle \hat{a}_{new} \rangle = 0$, which may be convenient for calculations. In particular when the pump beam has many photons, the quantum part (i.e. related to $\hat{a}_{new}$) may be neglected.

Additionally, $\hat{a}_{new}$ is an annihilation operator with the same anti-commutation relations as $\hat{a}_{old}$.

From mathematical point of view it is a perfectly legit operation. Moreover, both operators have the same domain, and spectrum only shifted by $\alpha_0$. To see that domain is the same take any $|\psi\rangle \in \text{dom}(\hat{a}_{old})$. Then denoting $|\phi\rangle := \hat{a}_{old} |\psi\rangle$, we check that $\hat{a}_{new} |\psi\rangle = |\phi\rangle-\alpha_0|\psi\rangle$ is a well-defined vector.

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