Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
The deficiency of $x$ is given by the formula $$D(x)=2x-\sigma(x).$$
A number $y$ is said to be deficient-perfect if $D(y) \mid y$.
Note that a deficient-perfect number $z$ is automatically deficient (i.e., $z$ satisfies $D(z) \geq 1$).
Here is my question:
If $a$ is deficient-perfect, can $ab$ be deficient-perfect if $b > 1$, and $ab$ is odd?
The reason for the condition $ab \equiv 1 \pmod 2$ is to exclude the case $$(a_1,b_1)=\bigg(2^k,2^{k+1}+2^t-1\bigg),$$ where $2^{k+1}+2^t-1$ is an odd prime and $t \leq k$, as well as the case $$(a_2,b_2)=(2^r,2^s).$$
MY ATTEMPT
Let $ab$ be odd. This implies that $a$ is odd and $b$ is odd.
Let $a$ be deficient-perfect, and suppose to the contrary that $ab$ is also deficient-perfect, where $b>1$.
Then we have the divisibility constraints $$D(a) \mid a \iff a = (2a - \sigma(a))c$$ $$D(ab) \mid ab \iff ab = (2ab - \sigma(ab))d$$
Substituting the first equation into the variable $a$ in the $\text{LHS}$ of the second equation, we obtain $$(2a - \sigma(a))bc = (2ab - \sigma(ab))d \iff 2abc - bc\sigma(a) = 2abd - d\sigma(ab)$$ $$\iff 2ab(c-d) = bc\sigma(a) - d\sigma(ab).$$
But $$c - d = \frac{a}{D(a)} - \frac{ab}{D(ab)} = a\cdot\bigg(\frac{D(ab)-bD(a)}{D(a)D(ab)}\bigg) = a\cdot\bigg(\frac{2ab-\sigma(ab)-2ab+b\sigma(a)}{D(a)D(ab)}\bigg)$$ $$=a\cdot\bigg(\frac{-\sigma(ab)+b\sigma(a)}{D(a)D(ab)}\bigg)$$ which implies that $c - d \leq 0$ since $$b\sigma(a) \leq \sigma(ab).$$
We therefore conclude that $$2ab(c-d) = bc\sigma(a) - d\sigma(ab) \leq 0.$$
Alas, this is where I get stuck.
Postscript: Note that $$\mathscr{D} = 9018009 = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}$$ is the lone odd deficient-perfect number that we know of, to this day.
So essentially, the question asks whether we could produce another odd deficient-perfect number from $\mathscr{D}$ (via multiplying $\mathscr{D}$ by $\mathscr{D'} > 1$), or if it could be proven that this is impossible.