Solution:
Given a subset $\left\{10^{a}+1,10^{b}+1,10^{c}+1\right\}$, we can directly expand the product of its elements:
$$\left(10^{a}+1\right)\left(10^{b}+1\right)\left(10^{c}+1\right)=10^{a+b+c}+10^{b+c}+10^{a+c}+10^{a+b}+10^{a}+10^{b}+10^{c}+1 .$$
In order for all digits to be 0 or 1, all 7 numbers $a+b+c, b+c, a+c, a+b, a, b, c$ should be distinct and nonzero, with the latter being guaranteed.
Without loss of generality, we can assume $c>b>a$. Then,
$$a+b+c>b+c>a+c>\max (a+b, c)$$
and
$$\min (a+b, c)>b>a$$
so the only two numbers that could be the same are $a+b$ and $c$.
There are 9 triples $(a, b, c)$ where $1 \leq a<b<c \leq 7$ and $a+b=c$, namely $(1,2,3),(1,3,4),(1,4,5)$, $(1,5,6),(1,6,7),(2,3,5),(2,4,6),(2,5,7)$, and $(3,4,7)$. The remaining triples $(a, b, c)$ all work, so the answer is $\binom{7}{3}-9=26$.