Given $n\in \mathbb{N}$, I would like to know why (preferably with an algebraic or analytic argument) $$\bigg(\Pi_{p\le n,\,p\, \text{prime}}(1-1/p)\bigg)\sum_{i}1/i=1,$$ where the sum is over all $i=2^{a_2}3^{a_3}\cdots p^{a_p}$ with $a_i\ge 0.$
This equation is due to the fact that $\bigg(\Pi_{p\le n,\,p\, \text{prime}}(1-1/p)\bigg)1/i=\mathbb{P}(X=i)$, where $X=\Pi_{p\le n}p^{Z_p}$, $Z_p$ are independent, geometric variables with $\mathbb{P}(Z_p=k)=(1-1/p)1/p^k$.
We have the power series expansion, valid for $|x| \lt 1$: $$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \ldots$$
which means that $$1 - \frac{1}{p} = \left(\frac{1}{1 - \frac{1}{p}}\right)^{-1} = \left(1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots\right)^{-1}.$$
We therefore have the product $$\prod\limits_{\substack{p \leq n \\ p\text{ prime}}}\left(1 - \frac{1}{p}\right) = \left(\prod\limits_{\substack{p \leq n \\ p\text{ prime}}}\left(1 + \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \ldots\right)\right)^{-1}.$$
The fundamental theorem of arithmetic says that every integer has a unique prime factorization, and so every positive integer divisible only by the primes not larger than $n$ will occur exactly once in the set $\{2^{a_2}3^{a_3} \ldots p^{a_p}: a_i \in \mathbb{N}\}$ (where $p$ is the largest prime $\leq n$).
But each of these terms also occurs exactly once in the denominators in the given product after using the Cauchy product, which can be done since the series are absolutely convergent and there are finitely many of them.
Intuitively, applying the distributive property, you will find the term $\frac{1}{2^{a_2}3^{a_3}\ldots p^{a_p}}$ occuring due to multiplying the $a_2$-th term of the series for $(1 + 1/2 + 1/2^2 + \ldots)$, the $a_3$-th term of the series for $(1 + 1/3 + 1/3^2 + \ldots)$, ..., the $a_p$-th term of the series for $(1 + 1/p + 1/p^2 + \ldots)$ (counting from $0$, so $\frac{1}{p_i^k}$ is the $k$-th term of the series summing over the $i$-th prime $p_i$).
That means $$\prod\limits_{\substack{p \leq n \\ p\text{ prime}}}\left(1 - \frac{1}{p}\right) = \left(\sum\limits_{a_i \in \mathbb{N}}\frac{1}{2^{a_2}3^{a_3}\ldots p^{a_p}}\right)^{-1}$$ which is what you observed.