This is extended Douglas Zare's comment explaining why we may take $|S|=1$. We prove that for any residue $\ell$ modulo $n$ there exist $a,b$ coprime to $n$ such that $\ell+a \equiv b \pmod n$. Then we may take $S=\{a/2\}$.

For any prime $p$ which divides $n$ choose remainder $a_p$ modulo $p$ such that neither $a_p$, nor $a_p+\ell$ is divisible by $p$. This is clearly possible, there are (at least) $p-2$ ways to do it. Then by Chinese Remainders Theorem there exists $a$ congruent to $a_p$ modulo each $p|n$. Use it.

As for the question 2, we may take ${\mathcal B}_n$ with $k$ elements, where $k$ is the number of distinct prime divisors of $n$ (this is in general better than powers of 2 not exceeding $n$). For any $p|n$ choose a number $b(p)$ such that $2b(p)$ gives remainder 2 modulo $p$ and 1 modulo other prime divisors of $n$. Fix $\ell$ and for $m=0,1,\dots$ denote by $f(m)$ the number of primes $p|n$ for which $p|\ell+m$.

**Lemma.** Assume that $f(m)\leqslant m$ and $f(m+1)\leqslant k-m$. Then there exists a set $A$, $|A|=m$, of prime divisors of $n$, such that $\ell+\sum_{p\in A} 2b(p)$ is coprime to $n$.

Proof. Denote $x=\ell+\sum_{p\in A} 2b(p)$. Then $x\equiv \ell+m+\chi_A(p) \pmod p$ for all prime $p|n$. So, there are $f(m)$ primes which have to be included to $A$ and $f(m+1)$ other primes which can not be included. Clearly such $A$ exists exactly when $f(m)\leqslant m$ and $f(m+1)\leqslant k-m$.

Now we need to find appropriate $m$. It is not a big deal, since we always have $f(m)+f(m+1)\leqslant k$ and $f(m)+f(m+2)\leqslant k$.

If $k$ is even, $k=2q$, we examine $m=q-1$ and $m=q$. Assume that both do not work. If $f(q)\leqslant q$, then $f(q-1)\geqslant q+1$, $f(q+1)\geqslant q+1$, a contradiction. If $f(q)\geqslant q+1$, then both $f(q+1)\leqslant q-1$ and $f(q+2)\leqslant q-1$ and we may take $m=q-1$.

If $k=2q-1$, consider $f(q)$. If $f(q)\leqslant q$, then $f(q-1)\geqslant q$ (else $m=q-1$ works) and $f(q+1)\geqslant q$ (else $m=q$ works). A contradiction. If $f(q)\geqslant q+1$, then $f(q+1)\leqslant q-2$, $f(q+2)\leqslant q-2$, so $m=q+1$ works.