Your condition is that $K$ be a kroneckerian field, namely either a totally real (as you mention) field or a totally imaginary quadratic extension of a totally real field: in this second case $K$ is said to be a CM field.

Since you have already treated in your question the case when $K$ is totally real, let me assume $K$ is CM. In this case, one has to notice that since a compositum of totally real fields is again totally real, we can speak of the maximal totally real subextension, call it $K^+$ of $K$. By assumption, $K^+\subsetneq K$ and since there is a totally real subfield $F\subseteq K$ such that $[K:F]=2$ we deduce $F=K^+$. Since you assumed $K$ to be Galois over $\mathbb{Q}$, I claim that $K^+/\mathbb{Q}$ is also Galois: this boils down to the statement that all embeddings $\iota:K^+\hookrightarrow\mathbb{R}$ have the same set-theoretic image, because this would imply that all roots of a minimal polynomial ok $K^+/\mathbb{Q}$ already lie in $K^+$. To check this, call $\mathbb{K}$ the subfield of $\mathbb{C}$ which is the image of one, and hence all by normality, embeddings of $K$ into $\mathbb{C}$: let $\mathbb{K}^+=\mathbb{K}\cap\mathbb{R}$. Then, for every $\iota:K^+\hookrightarrow\mathbb{C}$ we have $\iota(K^+)=\mathbb{K}^+$ since the image has to be contained in $\mathbb{R}\cap\mathbb{K}$ and needs be maximal. We have thus proven that $\langle\tau\rangle$ is a normal subgroup in $G$. This means that for each $g\in G$ we have $g\tau g^{-1}\in\langle\tau\rangle$ and since $\tau$ has order $2$ the only possibility is $g\tau g^{-1}=\tau$, namely $g\tau=\tau g$, proving that $\tau$ is central.

The other direction is obvious: if $\tau$ is central, then $\langle\tau\rangle$ is normal and if we set $K^+:=K^{\tau}$ we find a normal extension of $\mathbb{Q}$ (hence, either totally real or totally imaginary) that is fixed by at least one complex conjugation, hence is not totally imaginary and therefore is totally real. As $[K:K^+]=\#\langle\tau\rangle=2$, $K$ is a CM field.

Let me observe that in your very question you were implicitly assuming that speaking of "complex conjugation in $G$" is meaningful, but it can be shown that this is the case if and only if $K$ is (either totally real, or) a CM field.

Your condition is that $K$ be a kroneckerian field, namely either a totally real (as you mention) field or a totally imaginary quadratic extension of a totally real field: in this second case $K$ is said to be a CM field.

Since you have already treated in your question the case when $K$ is totally real, let me assume $K$ is CM. In this case, one has to notice that since a compositum of totally real fields is again totally real, we can speak of the maximal totally real subextension, call it $K^+$ of $K$. By assumption, $K^+\subsetneq K$ and since there is a totally real subfield $F\subseteq K$ such that $[K:F]=2$ we deduce $F=K^+$. Since you assumed $K$ to be Galois over $\mathbb{Q}$, I claim that $K^+/\mathbb{Q}$ is also Galois: this boils down to the statement that all embeddings $\iota:K^+\hookrightarrow\mathbb{R}$ have the same set-theoretic image, because this would imply that all roots of a minimal polynomial ok $K^+/\mathbb{Q}$ already lie in $K^+$. To check this, call $\mathbb{K}$ the subfield of $\mathbb{C}$ which is the image of one, and hence all by normality, embeddings of $K$ into $\mathbb{C}$: let $\mathbb{K}^+=\mathbb{K}\cap\mathbb{R}$. Then, for every $\iota:K^+\hookrightarrow\mathbb{C}$ we have $\iota(K^+)=\mathbb{K}^+$ since the image has to be contained in $\mathbb{R}\cap\mathbb{K}$ and needs be maximal. We have thus proven that $\langle\tau\rangle$ is a normal subgroup in $G$. This means that for each $g\in G$ we have $g\tau g^{-1}\in\langle\tau\rangle$ and since $\tau$ has order $2$ the only possibility is $g\tau g^{-1}=\tau$, namely $g\tau=\tau g$, proving that $\tau$ is central.

The other direction is obvious: if $\tau$ is central, then $\langle\tau\rangle$ is normal and if we set $K^+:=K^{\tau}$ we find a normal extension of $\mathbb{Q}$ (hence, either totally real or totally imaginary) that is fixed by at least one complex conjugation, hence is not totally imaginary and therefore is totally real. As $[K:K^+]=\#\langle\tau\rangle=2$, $K$ is a CM field.