This is surely a trivial question, since all sources seem to relegate it to an exercise. Here's my attempt at a proof. Is this the idea? Please excuse the wordiness of the exposition, it's from my own little collection of proofs I tex up along the way.
Let $J \subset \mathbb Z$ be a set of primes. Consider the spectrum \begin{equation*} S[J^{-1}] = \mathrm{hocolim}\left(S \xrightarrow{j_1} S \xrightarrow{j_1j_2} S \xrightarrow{j_1j_2j_3} \cdots\right). \end{equation*}
Since smash commutes with homotopy colimits (not sure if it does in general, but I'm pretty sure what we observe below is true) we get for any spectrum $X$ \begin{align*} \pi_*(S[J^{-1}] \wedge X) &= \pi_*\left( \mathrm{hocolim}\left(S \wedge X \xrightarrow{j_1} S \wedge X \xrightarrow{j_1j_2} S \wedge X \xrightarrow{j_1j_2j_3} \cdots\right) \right)\\ &\cong \mathrm{colim} \,\pi_\ast(X \wedge S)\\ &\cong \pi_*(X)[J^{-1}] \\ &\cong \pi_*(S)[J^{-1}] \otimes_{\pi_*(S)} \pi_*(X) \end{align*}
In what follows, we tensor over $\pi_*(S)$. We show that $X = S \wedge X \to S[J^{-1}] \wedge X$ induced by the obvious map on the first factors is an $S[J^{-1}]$ equivalence.
We consider the map \begin{equation} S[J^{-1}] \wedge S \wedge X \to S[J^{-1}] \wedge S[J^{-1}] \wedge X. \end{equation} Upon taking $\pi_\ast$, this can be identified with the map (do what we did above twice!) \begin{equation} \pi_\ast(S) \otimes \pi_*(X)[J^{-1}] \to \pi_\ast(S)[J^{-1}] \otimes \pi_*(X)[J^{-1}] \end{equation} given by $a \otimes b \mapsto a \otimes b$. This can further be identified with \begin{equation} \pi_*(X)[J^{-1}] \to \pi_\ast(S)[J^{-1}] \otimes \pi_*(X)[J^{-1}] \end{equation} via $b \mapsto 1 \otimes b$ which is an isomorphism since we have no $J$-torsion on both sides (In a low brow fashion, one can write out elementary tensors, adding factors of $\frac{j_i}{j_i}$ if necessary)
A remark: This is akin to the algebraic fact that \begin{equation} \mathbb Z \to \mathbb Z\left[\frac{1}{2}\right] \end{equation} induces an isomorphism \begin{equation} A \to \mathbb Z[J^{-1}]\otimes_{\mathbb Z} A \end{equation} if $A$ has no $J$-torsion.