Converting my comment to an answer.
It is wrong for infinite $S$. Consider the following partition of $\mathbb N=\lbrace 0,1,2,\dots\rbrace$. Define $P_1=4^1\mathbb N, P_2=4^2\mathbb N+1,P_3=4^3\mathbb N+3,P_4=4^4\mathbb N+5,P_5=4^5\mathbb N+6,\dots$ and more generally $P_k=4^k\mathbb N+m_k$ where $m_k$ is the smallest integer not in $P_1\cup\dots\cup P_{k-1}$ (with $m_1=0$.) By construction, the $P_k$ cover $\mathbb N$, and it remains to show that they don't overlap.
The sequence $m_k$ is easily seen to be increasing. If $P_a\cap P_b$ were non-empty for some $a, there would be non negative integers $r,s$ such that $m_a+ r4^a=m_b+s4^b=m_b+s4^{b-a}4^a$. Since $m_a, this would imply $r>s4^{b-a}$ and $m_b=m_a+(r-s4^{b-a})4^a\in P_a$ contradicting the definition of $m_b$. Thus $\mathbb N =P_1\sqcup P_2\sqcup P_3\sqcup\cdots$
However, $\sum_{i\geq 1}\frac{1}{4^i}=\frac{1}{4}\cdot\frac{1}{1-1/4}=\frac{1}{3}<1.$