With the helpful hints of others, I finally put together what I think is a full solution.
Reminder: $S = \mathbb{Z}^{+}$ and $S^{+} = S\cap (0,\infty)$ and $S^{-} = S\cap (-\infty,0)$.
$(\Rightarrow)$ I argue by contrapositive: Suppose $\sum_{n=1}^{\infty}|x_{n}|$ is not finite. Then
Claim: (from the hint) we have that at least one of $\sum_{n=1}^{\infty}f^{+}(x_{n})$ and $\sum_{n=1}^{\infty}f^{-}(x_{n})$ must be infinite, where
$f^{+}(x) = \begin{cases}x & x\geq 0\\ 0 & x < 0\end{cases}$
and $f^{-}(x) = \begin{cases}x & x < 0\\ 0 & x \geq 0\end{cases}$
(I assume this is what was meant by "padding" with $0$'s)
Indeed suppose they were both finite, then we could write
$\begin{eqnarray*}
\sum_{n=1}^{\infty}|x_{n}| &=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - f^{-}(x_{n})\\
&=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - \sum_{n=1}^{\infty}f^{-}(x_{n})\\
&=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - \sum_{n=1}^{\infty}f^{-}(x_{n})\\
\end{eqnarray*}$
which is the difference of two finite sums, this contradicts our original assumption and thus proves the claim.
Observation: Using this claim, we may assume without loss of generality that $\sum_{n=1}^{\infty}f^{+}(x_{n})$ is infinite.
Then for each $m\geq 1$, define the finite set $K_{m}:=\{1, \dots, m\}\cap S^{+}$, which is finite. Since $\sum_{n=1}^{\infty}f^{+}(x_{n})$ is infinite, the sequence
$$\sum_{n\in K_{m}}x_{n}$$ is unbounded as $m$ grows.
Now for the proof:
Now assume for a contradiction that $\sum_{n\in S}x_{n}$ converges unconditionally, say to $x\in\mathbb{R}$. Then (taking $\epsilon = 1$ in the definition of unconditional convergence) there exists a finite subset $J\subset S$ such that whenever $K\subset S$ is finite with $J\subset K$, we have
$$|\sum_{n\in K}x_{n} - x| < 1$$
By the triangle inequality, this provides the upper bound on $\sum_{n\in K}x_{n}$:
$$|\sum_{n\in K}x_{n}|\leq 1 + |x|$$
and this bound is satisfied so long as $J\subset K\subset S$ and $K$ is finite. Write $J^{+} = J\cap S^{+}$.
Apply this to the set $K_{m}\cup J$, for each $m\geq 1$, then we get
$\begin{eqnarray*}
|\sum_{n\in K_{m}\cup J}x_{n}| &=& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n} - \sum_{n\in K_{m}\cap J}x_{n}|\\
&\geq& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}| - |\sum_{n\in K_{m}\cap J}x_{n}|\\
&\geq& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}| - |\sum_{n\in J^{+}}x_{n}|\\
\end{eqnarray*}$
where the last line uses $J^{+}\supset K_{m}\cap J$ and the fact that all terms "over" both set are non-negative.
This contradicts the fact that $|\sum_{n\in K_{m}\cup J}x_{n}|$ is bounded, $|\sum_{n\in J^{+}}x_{n}|$ is a constant, and $|\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}|$ is unbounded.
$(\Leftarrow)$ Suppose $\sum_{n=1}^{\infty}|x_{n}|< \infty$. Then by completeness of the reals, there is $x\in\mathbb{R}$ such that $\sum_{n=1}^{N}x_{n} \to x$ as $N\to\infty$.
Let $\epsilon > 0$ be given.
Choose $N_{1}\geq 1$ so that $\sum_{n=m}^{\infty}|x_{n}|\leq \frac{\epsilon}{2}$ whenever $m\geq N_{1}$.
Choose $N_{2}\geq 1$ so that $|\sum_{n=1}^{m}x_{n} - x| <\frac{\epsilon}{2}$ whenever $m\geq N_{2}$.
Set $N = max(N_{1},N_{2})$, and set $J = \{1, ... , N\}$. Then whenever $K\subset S$ is finite such that $J\subset K$, we have
$\begin{eqnarray*}
|\sum_{n\in K}x_{n} - x| &=& |\sum_{n\in J}x_{n} - x + \sum_{n\in K\backslash J}x_{n}|\\
&\leq& |\sum_{n\in J}x_{n} - x| + |\sum_{n\in K\backslash J}x_{n}|\\
&\leq& |\sum_{n\in J}x_{n} - x| + \sum_{n\in K\backslash J}|x_{n}|\\
&\leq& \frac{\epsilon}{2} + \sum_{n = N + 1}^{\infty}|x_{n}|\\
&\leq& \epsilon
\end{eqnarray*}$
So the series converges to $x$ unconditionally.
I realize this could probably be done in a tighter fashion and is too long for me to ask anyone to check it over, but for now I'm just glad I see it. Thanks very much for the hints, and if there are any glaring problems that someone may find in my argument please tell me I would like to fix them.
