A function $f : \mathbb{N} \rightarrow \mathbb{Q}$ is said to be multiplicative if $$f(ab) = f(a)f(b)$$ whenever $\gcd(a,b)=1$.
It is known that the sum-of-divisors function $$\sigma(x) = \sum_{d \mid x}{d}$$ is multiplicative. It follows that the abundancy index function $$I(x) = \frac{\sigma(x)}{x}$$ is also multiplicative.
It is also known that the deficiency $$D(x) = 2x - \sigma(x)$$ and the sum-of-proper-divisors $$s(x) = \sigma(x) - x$$ functions are, in general, not multiplicative.
My question is:
Are there (restricted) instances when the deficiency and sum-of-proper-divisors functions are multiplicative?
MOTIVATION FOR THE QUESTION
When $yz$ is a perfect number for $\gcd(y,z)=1$, then we know that $$\sigma(yz) = \sigma(y)\sigma(z) = 2yz.$$
It turns out that we can also show that $$D(y)D(z) = 2s(y)s(z),$$ if $yz$ is a perfect number with $\gcd(y,z)=1$.
(Of course, I do not hope to show that $D(yz)=D(y)D(z)$ if $yz$ is perfect and $\gcd(y,z)=1$, since $yz$ is perfect implies that $D(yz)=0$. I just want to know if further simplified expressions may be obtained for either $D(y)D(z)$ or $s(y)s(z)$ (with $\gcd(y,z)=1$), whether or not $yz$ is perfect.)
None of these functions is multiplicative on any reasonable domain; in fact, for $x$ and $y$ coprime, we have $s(xy)=s(x)s(y)$ if and only if $x=y=1$, and $D(xy)=D(x)D(y)$ if and only if $\min\{x,y\}=1$.
Sufficiency is readily verified. For necessity, notice that $s(x)s(y)=s(xy)$ is equivalent to $(\sigma(x)−x)(\sigma(y)−y)=\sigma(xy)−xy$, which by multiplicativity of $\sigma$ is further equivalent to $yσ(x)+xσ(y)=2xy$, and eventually can be written as $$ \frac{\sigma(x)}x + \frac{\sigma(y)}y=2. $$ Taking into account that $\sigma(x)\ge x$ and $\sigma(y)\ge y$, with equalities if and only if $x=1$ and $y=1$, respectively, we conclude that $x=y=1$.
Similarly, for $x$ and $y$ coprime, the equality $D(xy)=D(x)D(y)$ can be equivalently rewritten as $$ (\sigma(x)-x)(\sigma(y)-y)=0, $$ which implies $x=1$ or $y=1$ since if we had $x,y>1$, then both factors in the LHS were strictly positive.