If $\mathcal{S}$ has only one element $p$, then $b_{i} = p^{i-1}$ and we can easily find $a_{1}, \ldots, a_{n}$ with $2 = \left\lceil \sum_{i=0}^{n-1} \frac{1}{p^{i}} \right\rceil = \sum_{i=0}^{n-1} \frac{a_{i}}{p^{i-1}}$ by taking $a_{1} = a_{2} = \cdots = a_{n-1} = 1$ and choosing $a_{n} = p^{n-1} - (p + p^{2} + \ldots + p^{n-2})$.

More generally, observe that the sum of $\frac{1}{b_{i}}$ over all $i$ is
$$\begin{aligned} \sum_{i} \frac{1}{b_{i}} &= \prod_{i} \left(1 + \frac{1}{p_{i}} + \frac{1}{p_{i}^{2}} + \ldots \right) \\ &= \prod_{p \in \mathcal{S}} \frac{p}{p-1} \end{aligned}$$
In particular, if $n$ is large enough, then
$$\left\lceil \sum_{j=1}^{n} \frac{1}{b_{j}} \right\rceil = \left\lceil \prod_{p \in \mathcal{S}} \frac{p}{p-1} \right\rceil$$
For the remainder of the proof, we will only consider $n$ large enough that this equality holds.

Next, we handle the special case $\mathcal{S} = \{2,3\}$, for which this product is $3$. Start by setting
$$a_{i} = \begin{cases}1, & \text{ if } 2 b_{i} \leqslant b_{n} \\ 2, & \text{ if } 2 b_{i} > b_{n}\end{cases}$$
Then,
$$\sum_{\substack{i \leqslant n \\ \nu_{3}(b_{i}) = t}} \frac{a_{i}}{b_{i}} = \begin{cases} \frac{2}{3^{t}}, & \text{ if } b_{n} \geqslant 3^{t} \\ 0, & \text{ otherwise } \end{cases}$$
As a result,
$$\begin{aligned} \sum_{i \leqslant n} \frac{a_{i}}{b_{i}} &= \sum_{\substack{t \geqslant 0 \\ 3^{t} \leqslant b_{n}}} \frac{2}{3^{t}} \\ &= 3 - \frac{1}{3^{T}} \end{aligned}$$
where $T$ is the largest $t \geqslant 0$ with $3^{t} \leqslant b_{n}$. Thus, increasing $a_{j}$ by one (where $b_{j} = 3^{T}$) gives a sequence of $a_{i}$ that works.

Otherwise, we may assume that $|\mathcal{S}| > 1$ and $\mathcal{S} \neq \{2,3\}$, which means that the product $\prod_{p \in \mathcal{S}} \frac{p}{p-1}$ is not an integer. Indeed,
- if $|\mathcal{S}| > 2$ then $2$ divides the denominator at least twice and so divides the denominator of the overall fraction;
- if $|\mathcal{S}| = 2$ and $2 \notin \mathcal{S}$ then $2$ divides the denominator and not the numerator;
- if $\mathcal{S} = \{2, p\}$ then the product is $2p/(p-1)$ which is not an integer for $p > 3$.

It follows that for some fixed $\alpha > 0$, we have that
$$\left\lceil \prod_{p \in \mathcal{S}} \frac{p}{p-1} \right\rceil = \prod_{p \in \mathcal{S}} \frac{p}{p-1} + \alpha$$
from which it follows that
$$\left\lceil \sum_{i=1}^{n} \frac{1}{b_{i}} \right\rceil - \sum_{i=1}^{n} \frac{1}{b_{i}} > \alpha$$
It will now suffice to prove the following claim.

Claim. Suppose that $n$ is large enough, and let $e_{p}$ be the largest nonnegative integer such that $p^{e_{p}} \leqslant b_{n}$. Let $M = \prod_{p \in \mathcal{S}} p^{e_{p}}$. If $u$ is a positive integer such that $u / M > \alpha$, then there exist nonnegative integers $a_{i}$ such that
$$\sum_{i} \frac{a_{i}}{b_{i}} = \frac{u}{M}.$$
The problem statement follows after replacing $a_{i}$ with $a_{i} + 1$ for each $i$.

To prove this, choose some constant $c$ such that $\sum_{p \in \mathcal{S}} p^{-c} < \alpha$, and suppose $n$ is large enough that $p^{c} < b_{n}$ for each $p \in \mathcal{S}$; in particular, $p^{c} \mid M$ with $M$ defined as above.

For each $p \in \mathcal{S}$, let $i_{p}$ be such that $b_{i_{p}} = p^{e_{p}}$ and choose the smallest nonnegative integer $a_{i_{p}}$ satisfying
$$p^{e_{p}-c} \left\lvert\, a_{i_{p}} \left( \frac{M}{p^{e_{p}}} \right) - u \right.$$
Such an $a_{i_{p}}$ must exist and be at most $p^{e_{p}-c}$; indeed, $\frac{M}{p^{e_{p}}}$ is an integer coprime to $p$, so we can take $a_{i_{p}}$ to be equal to $u$ times its multiplicative inverse modulo $p^{e_{p}-c}$.

The sum of the contributions to the sum from the $a_{i_{p}}$ is at most
$$\sum_{p \in \mathcal{S}} \frac{p^{e_{p}-c}}{p^{e_{p}}} = \sum_{p \in \mathcal{S}} p^{-c} < \alpha$$
So, we have
$$\frac{u}{M} = \sum_{p \in \mathcal{S}} \frac{a_{i_{p}}}{p^{e_{p}}} + \frac{r}{\prod_{p \in \mathcal{S}} p^{c}},$$
where $r$ is an integer because of our choice of $a_{i_{p}}$ and $r$ is nonnegative because of the bound on $u$. Simply choose $a_{i} = r$ where $b_{i} = \prod_{p \in \mathcal{S}} p^{c}$ to complete the proof.

We reduce to the claim as in Solution 1, and provide an alternative approach for constructing the $a_{i}$.

Let $p_{0} \in \mathcal{S}$ be the smallest prime in $\mathcal{S}$. Let $z_{0} = u / M$. We construct a sequence $z_{0}, z_{1}, z_{2}, \ldots$ and values of $a_{i}$ by the following iterative process: to construct $z_{j+1}$,
- select the largest prime $p \in \mathcal{S}$ dividing the denominator of $z_{j}$, and let $\mu$ be the number of times $p$ divides the denominator of $z_{j}$;
- choose the largest $\nu$ such that $p_{0}^{\nu} p^{\mu} \leqslant b_{n}$, and let $i \leqslant n$ be such that $b_{i} = p_{0}^{\nu} p^{\mu}$;
- choose $0 \leqslant a_{i} < p$ such that the denominator of $z_{k} - a_{i} / b_{i}$ has at most $\mu - 1$ factors of $p$, and let $z_{k+1} = z_{k} - a_{i} / b_{i}$;
- continue until $p_{0}$ is the only prime dividing the denominator of $z_{k}$.

Note that we can always choose $a_{i}$ in step 3; by construction, $z_{k} b_{i}$ has no factors of $p$ in its denominator, so must be realised as an element of $\mathbb{Z}_{p}$.

Each time we do this, $b_{i} > M / p_{0}$ by construction, so
$$\frac{a_{i}}{b_{i}} < \frac{p p_{0}}{M} \leqslant \frac{p_{0} p_{1}}{M},$$
where $p_{1}$ is the largest prime in $\mathcal{S}$.
And the number of times we do this operation is at most
$$\sum_{\substack{p \in \mathcal{S} \\ p > p_{0}}} e_{p} \leqslant |\mathcal{S}| \log_{2}(M)$$
so the sum of the $a_{i} / b_{i}$ we have assigned is at most $|\mathcal{S}| p_{0} p_{1} \log_{2}(M) / M$.

Choose $n$ large enough that $\log_{2}(M) / M < \alpha$; after subtracting the above choices of $a_{i} / b_{i}$ from $u / M$, we have a quantity of the form $r / p_{0}^{e_{p_{0}}}$, where $r$ is an integer by construction and $r$ is positive by the above bounds. Simply set $a_{i} = r$ where $b_{i} = p_{0}^{e_{P_{0}}}$ to complete the proof.

As in Solution 1, we may handle $|\mathcal{S}| = 1$ and $\mathcal{S} = \{2,3\}$ separately; otherwise, we can define $\alpha$ as we did in that solution. Also define $e_{p}$ to be the largest nonnegative integer such that $p^{e_{p}} \leqslant b_{n}$ as we did in Solution 1.

We will show that, for $n$ sufficiently large, we may choose some $j \leqslant n$, and positive integers $a_{i}$, such that
$$\sum_{i \neq j} \frac{a_{i}}{b_{i}} - \sum_{i \neq j} \frac{1}{b_{i}} < \alpha$$
and all $\frac{a_{i}}{b_{i}}$ are integer multiples of $\frac{1}{b_{j}}$. We then set $a_{j}$ to be the least positive integer such that the sum on the left is an integer, which will obviously have the required value.

Concretely, choose $j$ such that $b_{j} = \prod_{p \in \mathcal{S}} p^{\left.e_{p} / |\mathcal{S}| \right]}$, which is less than $b_{n}$ by construction. For $i \neq j$, set $a_{i} = b_{i} / \operatorname{gcd}(b_{i}, b_{j})$. We have
$$\sum_{i \neq j} \frac{a_{i}}{b_{i}} - \sum_{i \neq j} \frac{1}{b_{i}} < \sum_{\substack{i \neq j \\ a_{i} > 1}} \frac{a_{i}}{b_{i}}$$
If $a_{i} > 1$, then there must be some $p \in \mathcal{S}$ for which $p^{\left\lfloor e_{p} / |\mathcal{S}| \right] + 1} \mid b_{i}$, and so
$$\frac{a_{i}}{b_{i}} = \frac{1}{\operatorname{gcd}(b_{i}, b_{j})} \leqslant \frac{1}{p^{\left[e_{p} / |\mathcal{S}| \right]}} < \frac{p}{b_{n}^{1 / |\mathcal{S}|}}$$
where the last inequality follows from the fact that $p^{e_{p} + 1} > b_{n}$.

Now $n \leqslant \prod_{p \in \mathcal{S}} (\log_{p}(b_{n}) + 1) \leqslant (2 \log b_{n})^{|\mathcal{S}|}$, so
$$\sum_{\substack{i \neq j \\ a_{i} > 1}} \frac{a_{i}}{b_{i}} \leqslant \frac{(2 \log b_{n})^{|\mathcal{S}|}}{b_{n}^{1 / |\mathcal{S}|}}$$
and so we can choose $n$ large enough that this quantity is less than $\alpha$, as required.