For the first question, yes, $S \setminus A$ is contractible. Rewrite the standard basis so that it is indexed by $\mathbb{Z}$. Let $U$ be the bilateral shift. Then $U$ is an isomorphism on $\ell^p$ for any $0 < p \leq 2$. Recall, from this paper, that $\varphi(x) = \frac{1}{2}(1 - \|x\|)e_0 + U(x)$ is a homeomorphism from the unit ball of $\ell^2$ onto itself without fixed point. From the geometric description of $\varphi$, contained in the proof of Theorem 1 of the cited paper, and using the facts that $e_0 \in \ell^p$ and $U$ is an isomorphism of $\ell^p$ for all $0 < p < 2$, it is not hard to see that $\varphi$ restricts to a homeomorphism from the unit ball of $\bigcup_{0 < p < 2} \ell^p$ (under the $\ell^2$ norm) to itself. Then $\psi$, which sends $x$ to the point at which the ray from $\varphi(x)$ to $x$ meets the unit sphere, is a retraction from the unit ball of $\bigcup_{0 < p < 2} \ell^p$ onto the unit sphere of $\bigcup_{0 < p < 2} \ell^p$ (i.e., $S \setminus A$). Thus, $S \setminus A$ is contractible, as $\Phi(x, t) = \psi(tx)$ is a homotopy between a constant map and the identity map on $S \setminus A$.
On second thought, I decided to still write down my approach to proving $A$ is path-connected despite Aleksei's wonderful (and simpler) answer, to ensure there is one complete answer.
For any $x \in A$, denote $x_e$ to be the part of $x$ supported on even indices and $x_o$ to be the part of $x$ supported on odd indices. Let $x, y \in A$. We first show that there is a path from $x$ to $x'$ in $A$, and from $y$ to $y'$ in $A$, s.t. $x'$ and $y'$ have disjoint supports.
Since $x \notin \bigcup_{0 < p < 2} \ell^p$, we must have at least one of $x_e$ and $x_o$ is not in $\bigcup_{0 < p < 2} \ell^p$ either. The same holds for $y$. If $x_e \notin \bigcup_{0 < p < 2} \ell^p$ and $y_o \notin \bigcup_{0 < p < 2} \ell^p$, then simply let $x' = \frac{x_e}{\|x_e\|}$. Now define,
$$f(t)_n = \begin{cases}
x_n &, \text{ if }n\text{ is even}\\
(1 - t)x_n &, \text{ if }n\text{ is odd}
\end{cases}$$
$$g(t) = \frac{f(t)}{\|f(t)\|}$$
It is then easy to verify that $g$ is a path from $x$ to $x'$ in $A$. Similarly, there is a path from $y$ to $y' = \frac{y_o}{\|y_o\|}$ in $A$. Clearly, $x'$ and $y'$ have disjoint supports. Similar arguments also apply when $x_o \notin \bigcup_{0 < p < 2} \ell^p$ and $y_e \notin \bigcup_{0 < p < 2} \ell^p$.
Now, consider when $x_o \notin \bigcup_{0 < p < 2} \ell^p$ and $y_o \notin \bigcup_{0 < p < 2} \ell^p$. The point here is we can first rotate $x$ to $x''$ s.t. $(x'')_e \notin \bigcup_{0 < p < 2} \ell^p$, then apply the result of the previous paragraph. This can be easily achieved by,
$$f(t)_n = \begin{cases}
\cos(\frac{\pi}{2}t)x_n + \sin(\frac{\pi}{2}t)x_{n+1} &, \text{ if }n\text{ is odd}\\
-\sin(\frac{\pi}{2}t)x_{n-1} + \cos(\frac{\pi}{2}t)x_n &, \text{ if }n\text{ is even}
\end{cases}$$
Then set $x'' = f(1)$. Since $(x'')_e$ is simply $x_o$ shifted one index forward, we have $(x'')_e \notin \bigcup_{0 < p < 2} \ell^p$. $f(t)$ is simply rotation in each pair of adjacent coordinates, so it preserves the $\ell^2$ norm. Any rotation matrix is also both bounded below and above under $\ell^p$ norm for any $0 < p < 2$, so as $x \notin \bigcup_{0 < p < 2} \ell^p$, $f(t) \notin \bigcup_{0 < p < 2} \ell^p$ either for any $t$. Thus, $f(t)$ is indeed a path in $A$. Now apply the result in the previous paragraph to reach the desired conclusion. The case where $x_e \notin \bigcup_{0 < p < 2} \ell^p$ and $y_e \notin \bigcup_{0 < p < 2} \ell^p$ can be proved similarly.
Now, it suffices to show that for $x, y \in A$ with disjoint supports, they can be connected by a path. Such a path is simply given by,
$$f(t) = \frac{tx + (1 - t)y}{\|tx + (1 - t)y\|}$$
