Let $K$ be the maximal abelian extension of $\mathbb{Q}$ that that $\operatorname{Gal}(K/\mathbb{Q})$ has exponent $2$ and is unramified at all finite odd primes.
Claim: $K = \mathbb{Q}(\sqrt{2}, i)$.
Proof: Let $F$ be any finite subextension of $K$. Then $F/\mathbb{Q}$ is abelian of exponent $2$, so $[F : \mathbb{Q}] = 2^m$ for some $m$. Chapter 4, Exercise 32 of 'Number Fields' by Marcus shows that $F \subseteq \mathbb{Q}(\zeta_{2^m})$, where $\zeta_{2^m}$ is a primitive $2^m$th root of unity for some $m$. Since the finite subextension $F$ was arbitrary, it follows that $$ K \subseteq \mathbb{Q}(\zeta_{2^\infty}) := \varprojlim_m \mathbb{Q}(\zeta_{2^m}). $$ Let $G = \operatorname{Gal}(\mathbb{Q}(\zeta_{2^\infty}) / \mathbb{Q}) \cong \mathbb{Z}_2^\times$, where $\mathbb{Z}_2$ denotes the $2$-adic integers. Then $K$ is the fixed field of the maximal subgroup $H\leq G$ such that $G/H$ has exponent $2$. Clearly $\mathbb{Z}_2^{\times 2} \subseteq H$. Since $\mathbb{Z}_2^\times / \mathbb{Z}_2^{\times 2}$ is $2$-torsion, it follows that $H = \mathbb{Z}^{\times 2}$, so $[K:\mathbb{Q}] = 4$. It is easy to see that $\mathbb{Q}(\sqrt{2},i)\subseteq K$, so the result follows.
$$\tag*{$\blacksquare$}$$
My Question: The proof above is a bit ad-hoc. I suspect there is a more systematic way to do this using idelic class field theory. Could someone provide a slick proof using the idelic language?