Question
Is the set of almost perfect numbers closed under multiplication?
It is known that $2^k$ for integral $k \geq 0$ are almost perfect. It is also known (by work of Antalan and Tagle) that if $M \neq 2^k$ is an even almost perfect number, then $M = 2^r b^2$ where $b$ is an odd composite. Finally, it is known (by work of Antalan and Dris) that $b^2$ is not almost perfect.
Hence, it seems easy to show the following Proposition.
Proposition The set APN of almost perfect numbers is closed under multiplication if and only if APN $= \left\{2^k | k \in \mathbb{Z}^+ \cup \{0\}\right\}$.
This is too long to be posted in the Comments section and is not really an answer - I just wanted to collect some of my recent thoughts on this problem here.
The definition of a set $S$ being closed under some binary operation $*$ is the following: $$\forall a, b \in S, a * b \in S.$$
The negation of this statement is: $$\exists a, b \in S, a * b \not\in S.$$
So now suppose that the set of almost perfect numbers APN is not closed under multiplication.
Then there exists $x, y \in$ APN, such that $xy \not\in$ APN.
But $x, y \in$ APN means that $$\sigma(x) = 2x - 1$$ and $$\sigma(y) = 2y - 1$$ where $\sigma = \sigma_{1}$ is the classical sum-of-divisors function.
This implies that $$2xy - D(xy) = \sigma(xy) \leq \sigma(x)\sigma(y) = (2x - 1)(2y - 1) = 4xy - 2x - 2y + 1$$ from which it follows that $$D(xy) \geq -2xy-2x-2y+1 = 3-2(x+1)(y+1),$$ where $D(n)=2n-\sigma(n)$ is the deficiency of $n \in \mathbb{N}$, and $D(xy) \neq 1$ (since $xy \not\in$ APN).
Alas, we do not arrive at a contradiction.