The answer is **no**. Below I give a proof for the $n=1$ case that the only operators obeying these constraints are of the form
$$B=b P_1+ b^{-1} P_2 \quad , \tag 1$$
where $P_1=\sum\limits_{n=0}^\infty |2n\rangle\langle 2n|$ and $P_2=\sum\limits_{n=0}^\infty |2n+1\rangle\langle 2n+1|$ for any $b\in \mathbb R\setminus\{0\}$. 

It is easy to see that $B$ is hermitian and that $B=\pm I$ if and only if $b=\pm1$. As a proof that $B$ in $(1)$ obeys the requirements, we compute:

$$BaB=\left(bP_1+b^{-1}P_2\right) a\left(bP_1+b^{-1}P_2\right) = P_1 aP_2 + P_2 aP_1 \quad , \tag 2 $$

where we've used the fact that $P_1aP_1 = P_2aP_2 =0$. Note that the RHS of $(2)$ is indeed independent of $b$. It remains to show that $a=P_1aP_2 + P_2 aP_1$. To this end, we compute in a straightforward manner for all $n,m\in \mathbb N_0$ the matrix elements:

$$A_{nm}:=\langle n|P_1 aP_2 + P_2 a P_1|m\rangle = \langle n|P_1 aP_2|m\rangle  + \langle n|P_2aP_1|m\rangle \quad . \tag 3$$

We see that if both $n$ and $m$ are either even or odd, the matrix elements vanish, so $A_{nm}=0+0 =\langle n|a|m\rangle=0$. Next, we consider $n$ even and $m$ odd and get $A_{nm}= \langle n|a|m\rangle +0$ and when $n$ odd and $m$ even, then $A_{nm}=0 0 + \langle n|a|m\rangle$. In conclusion, we have shown $A_{nm}=\langle n|a|m\rangle$ for all $n,m \in \mathbb N_0$ and hence $A=a$, which finally shows $BaB=a$ with $B$ defined in $(1)$. 

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Let us now give a proof that operators $B$ obeying $BaB=a$ are necessarily of the form $(1)$. 

We start by recalling the [fact](https://physics.stackexchange.com/questions/155852/is-there-a-simple-way-of-finding-the-eigenstates-of-the-creation-and-annihilatio) that $a^\dagger$ has no eigenvectors. In particular, for every eigenvector $|b\rangle$ of $B$ it holds that $a^\dagger |b\rangle \neq 0$ and thus, by assumption: 

$$0 \neq a^\dagger |b\rangle = Ba^\dagger B|b\rangle = b Ba^\dagger |b\rangle \quad . \tag 4$$ 

This shows that $b\neq 0$ and $Ba^\dagger |b\rangle\neq 0$, and thus that $a^\dagger |b\rangle$ is an eigenstate of $B$ with eigenvalue $1/b$; in particular $B$ has no vanishing eigenvalue.

Next, the assumptions also imply that

$$a|0\rangle = BaB|0\rangle = 0\tag 5 \quad , $$

which, together with the previous considerations show that $aB|0\rangle=0$ (because else it would be an eigenstate of $B$ with eigenvalue $0$), and since the vacuum/non-degeneracy of the ground state of the harmonic oscillator is unique, we must have that $B|0\rangle= b|0\rangle$ for some (non-zero) eigenvalue $b$. 

Finally, consider the state $|1\rangle=a^\dagger |0\rangle$. From $(1)$ we know that $B|1\rangle=Ba^\dagger |0\rangle=1/b |1\rangle$. Similarly, we know $B|2\rangle=\frac{1}{\sqrt 2}Ba^\dagger|1\rangle = b|2\rangle$. The pattern should be clear now: We can re-write $B$ in the occupation-number basis as
$$ B = b P_1 +b^{-1} P_2\tag 6 \quad ,$$  

where $P_1$ projects on the even and $P_2$ on the odd subspace of number states, as defined above.