The answer to your amended question is yes. In fact, for any $\epsilon\in[0,1)$ we have
$$
\mathbb{P}(\vert S\vert > \epsilon)\ge (1-\epsilon^2)^2/3.
$$
So, we can take $\delta = 1-(1-\epsilon^2)^2/3$. This is the $L^0$ version of the Khintchine inequality.

To prove it, you can use $\mathbb{E}[X_iX_j^3]=0$ for $i\not=j$ and $X_i^4=X_i^2X_j^2=1$ to get
$$
\begin{align}
\mathbb{E}[S^4]&=\sum_ia_i^4+3\sum_{i\not=j}a_i^2a_j^2=3\left(\sum_ia_i^2\right)^2-2\sum_ia_i^4\\\\
&\le 3.
\end{align}
$$
The Paley-Zygmund inequality gives
$$
\begin{align}
\mathbb{P}(\vert S\vert >\epsilon)&\ge(1-\epsilon^2)^2\frac{\mathbb{E}[S^2]^2}{\mathbb{E}[S^4]}\\\\
&\ge(1-\epsilon^2)^2/3.
\end{align}
$$

This bound gives $\delta=2/3$ for $\epsilon=0$. By considering the example with $a_1=a_2=1/\sqrt{2}$ and $a_i=0$ for $i > 2$, which satisfies $\mathbb{P}(S=0)=1/2$ we see that it is necessary that $\delta\ge1/2$. In fact, a simple argument noting that the distribution is symmetric under a sign change for $X_1$ (as mentioned by Luca in the comments) shows that $\mathbb{P}(S=0)\le1/2$.

See also the paper On Khintchine inequalities with a weight, where they prove the same bound as I just did above. Also, using the optimal constants for the $L^p$ Khintchine inequality, as in Lemma 3 of that paper, gives an improved bound for $\mathbb{P}(\vert S\vert\le\epsilon)$ tending to $1-2e^{-2+\gamma}\approx0.517$ as $\epsilon$ goes to zero, which is close to optimal.

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