Let $\mathbf A_T$ denote the restricted direct product $\mathbf R\times \prod'_{p\in T}\mathbf Q_p$ (relative to the subgroups $\mathbf Z_p$, $p\in T$).

The OP asked whether the Pontryagin dual of $S^{-1}\mathbf Z$ is isomorphic to $\mathbf A_T/(S^{-1}\mathbf Z)$. Let's take it up in two steps:

Step 1: $\mathbf Q_p/\mathbf Z_p$ is canonically isomorphic (as an abstract group) to the subgroup of $\mathbf R/\mathbf Z$ consisting of elements whose order is a power of $p$ (mapping $p^{-n}\in \mathbf Q_p$ to $p^{-n}\in \mathbf R/\mathbf Z$ determines this embedding). This gives rise to an element $\psi_p\in \hat{\mathbf Q}_p$ which vanishes on $\mathbf Z_p$ but not on $p^{-1}\mathbf Z_p$. Also define $\psi_\infty$ to be the quotient map $\mathbf R\to \mathbf R/\mathbf Q$ composed with the inversion map $x\mapsto -x$.

Step 2: The formula $(a_p)\mapsto ((x_p)\mapsto \exp(2\pi i \sum_{p\in T\cup \{\infty\}} \psi_p(a_p x_p)))$ defines an isomorphism $\mathbf A_T\to \widehat{\mathbf A_T}$.
Under this isomorphism $S^{-1}\mathbf Z$ maps to precisely those characters which vanish on $S^{-1}\mathbf Z$, establishing the isomorphism $S^{-1}\mathbf Z=\widehat{\mathbf A_T/S^{-1}\mathbf Z}$ (this can be seen with the help of partial fraction expansions of rational numbers).