Hints to the question (v1):

1. Recall that the operator time ordering is symmetric $$\tag{1}T[A_1(t_1)\ldots A_n(t_n)]~=~T[A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)})], $$ where $\pi\in S_n$ is a permutation. (Here we assume for simplicity that all operators are Grassmann-even.)

2. Recall that if $t_1< \ldots <t_n$, then the operator time ordering is defined $$\tag{2}T[A_n(t_n)\ldots A_1(t_1)]~:=~A_n(t_n)\ldots A_1(t_1).$$

3. It becomes a bit tedious to explain the time ordering rule $T[A_n(t_n)\ldots A_1(t_1)]$ when a subset of the times $t_1,\ldots, t_n$, happens to be exactly equal. Of course the operators $A_n(t_n)\ldots A_1(t_1)$ should be symmetrized in an appropriate sense.

4. To avoid that technical point 3, let us discretize time. More precisely, let us assume that the three different operators $H(t)$, $A(t)$, and $B(t)$ live on three different time discretizations, such that two operators of different type are never taken at exactly the same instant. Then that we can easily time-order any operator expression only knowing the un-equal-time ordering rules (1)-(2).

5. In OP's sought-for identity, replace time integrations with appropriate discrete sums of operators that lives on their respective sub-lattices. It turns out that the corresponding discretized version of OP's sought-for identity is simple to verify.

6. At the end of the calculation, take the continuum limit where the lattice constant goes to zero, and summations becomes integrals again. Argue that the identity continues to hold.

Hints to the question (v1):

1. Recall that the operator time ordering is symmetric $$\tag{1}T[A_1(t_1)\ldots A_n(t_n)]~=~T[A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)})], $$ where $\pi\in S_n$ is a permutation. (Here we assume for simplicity that all operators are Grassmann-even. Else there will be additional sign factors.)

2. Recall that if $t_1> \ldots >t_n$, then the operator time ordering is defined $$\tag{2}T[A_1(t_1)\ldots A_n(t_n)]~:=~A_1(t_1)\ldots A_n(t_n).$$

3. It becomes a bit technical to explain and work with the time ordering rule $T[A_1(t_1)\ldots A_n(t_n)]$ when a subset of the times $t_1,\ldots, t_n$, happens to be exactly equal. Of course in this case, the operators $A_1(t_1)\ldots A_n(t_n)$ should be symmetrized in an appropriate sense.

4. Extend the definition of time-ordering $T[A_1(t_1)\ldots A_n(t_n)]$ by multi-linearity.

5. To avoid that technical point 3, let us discretize time. More precisely, let us assume that OP's three different operators $H(t)$, $A(t)$, and $B(t)$ live on three different time discretizations, such that two different operators are never taken at exactly the same instant. (That e.g. a power of $H(t)$ appears in the same time $t$ does not matter, since $H(t)$ commutes with itself, so we can ignore the symmetrization procedure from point 3 without introducing errors.) In that way we can easily time-order any operator expression of $H(t)$, $A(t)$, and $B(t)$, only knowing the un-equal-time ordering rules (1)-(2).

6. In OP's sought-for identity, replace time integrations $\int\! dt$ with appropriate discrete sums $\sum$ of operators that live on their respective sub-lattices of time. The corresponding discretized version of OP's sought-for identity becomes a trivial identity, since the LHS is defined as the RHS.

7. At the end of the calculation, take the continuum limit where the lattice constant goes to zero, and summations becomes integrals again. Argue that the identity continues to hold.