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I do not know if this is helpful but the question is clearly related to the Shafarevich-Tate group of some algebraic torus. Namely, let's consider algebraic torus $T=Res_{K/\mathbb Q}\mathbb G_m/\mathbb G_m$, where $Res_{K/\mathbb Q}\mathbb G_m$ is the Weil restriction of $\mathbb G_m$ from $K$ to $\mathbb Q$. Then from Hilbert 90 $T(\mathbb Q)=K^\times/\mathbb Q^\times$. Consider the short exact sequence $0\rightarrow T[2] \rightarrow T \xrightarrow{\times 2} T \rightarrow 0$, and the corresponding exact sequence of cohomology groups gives an embedding of $$(K^\times/\mathbb Q^\times)/(K^\times/\mathbb Q^\times)^{\times 2}\simeq K^\times/(K^{\times 2}\cdot \mathbb Q^\times) \hookrightarrow H^1(\mathrm{Gal(\overline{\mathbb Q}/\mathbb Q)}, T[2](\overline{\mathbb Q})) $$ And for every $p$, the points $T(\mathbb Q_p)=(\prod_{\mathfrak p|p}K_{\mathfrak p}^\times)/\mathbb Q_p^\times$ and so $$ (\prod_{\mathfrak p|p}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times 2})/ \mathbb Q_p^\times\hookrightarrow H^1(\mathrm{Gal(\overline{\mathbb Q_p}/\mathbb Q_p)}, T[2](\overline{\mathbb Q_p})) $$ So if you want to prove that $\alpha \in K^\times/(K^{\times 2}\cdot \mathbb Q^\times)$ is zero if and only if it is zero in $(\prod_{\mathfrak p|p}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times 2})/ \mathbb Q_p^\times$ for each $p$ you need to prove that the image of $K^\times/(K^{\times 2}\cdot \mathbb Q^\times)$ in $$ \underline{Ш}^1(T[2])= \mathrm{Ker}(H^1(\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q,T[2](\overline{\mathbb Q}))\rightarrow \bigoplus _{p}H^1(\mathrm{Gal(\overline{\mathbb Q_p}/\mathbb Q_p)}, T[2](\overline{\mathbb Q_p}))) $$ is zero. In particular this is true if $\underline{Ш}^1(T[2])$ is 0.