Suppose that you have a collection of integers $f_{\iota}$ indexed by embeddings $\iota:K\hookrightarrow\mathbb{C}$ and satisfying for some integer $m$ (this is slightly different from the OP's identity): $$\prod_{\iota}\iota(\epsilon)^{mf_{\iota}}=1,\text{ for all $\epsilon\in\mathcal{O}_K^\times$}.$$

To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ is independent of $c$ and $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have: $$ f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}. $$ Another way to write this is, for all $\iota$: $$ \sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}. $$

Let us return to the first identity above. Taking the logarithm of its absolute value gives us: ` $$ \sum_{\iota}f_{\iota}\vert \iota(\epsilon)\vert=0. $$

But consider the map $$ \ell:\mathcal{O}_K^\times\otimes\mathbb{R}\rightarrow\oplus_{v\vert\infty}\mathbb{R}. $$given by $\ell(\epsilon\otimes 1)_v=\log\vert \epsilon\vert_v$ if $v$ is real and by $\ell(\epsilon\otimes 1)_v=2\log\vert \epsilon\vert_v$ if $v$ is complex. The indexing set here is the set of inequivalent archimedean norms on $K$. Then Dirichlet's unit theorem says that $\ell$ is an isomorphism onto the hyperplane $H$ where the sum of the co-ordinates is identically $0$.

Set $w(\sigma)=f_{\sigma\iota}+f_{\sigma\bar{\iota}}$: this doesn't depend on $\iota$. To finish the proof, we must show that it is independent of $\sigma$ as well. First, let $L$ be as above. From what we have proven already, we find that $L$ is stable under all complex conjugations. Assume that $L$ is totally complex; this implies that $K$ is also totally complex. Then, if $n=[K:\mathbb{Q}]$, we have: $$ 2nw(1)=\sum_{\iota}(f_{\iota}+f_{\bar{\iota}})=\sum_{\iota}(f_{\sigma\iota}+f_{\sigma\bar{\iota}})=2nw(\sigma). $$ So, in this case, the independence is clear.

Suppose that you have a collection of integers $f_{\iota}$ indexed by embeddings $\iota:K\hookrightarrow\mathbb{C}$ and satisfying for some integer $m$ (this is slightly different from the OP's identity):

$$\prod_{\iota}\iota(\epsilon)^{mf_{\iota}}=1,\text{ for all $\epsilon\in\mathcal{O}_K^\times$}.$$

To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ is independent of $c$ and $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have:

$$ f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}. $$

Another way to write this is, for all $\iota$:

$$ \sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}. $$

Let us return to the first identity above. Taking the logarithm of its absolute value gives us: 

$$ \sum_{\iota}f_{\iota}\vert \iota(\epsilon)\vert=0. $$

But consider the map

$$ \ell:\mathcal{O}_K^\times\otimes\mathbb{R}\rightarrow\oplus_{v\vert\infty}\mathbb{R}. $$ given by $\ell(\epsilon\otimes 1)_v=\log\vert \epsilon\vert_v$ if $v$ is real and by $\ell(\epsilon\otimes 1)_v=2\log\vert \epsilon\vert_v$ if $v$ is complex. The indexing set here is the set of inequivalent archimedean norms on $K$. Then Dirichlet's unit theorem says that $\ell$ is an isomorphism onto the hyperplane $H$ where the sum of the co-ordinates is identically $0$.

Set $w(\sigma)=f_{\sigma\iota}+f_{\sigma\bar{\iota}}$: this doesn't depend on $\iota$. To finish the proof, we must show that it is independent of $\sigma$ as well. First, let $L$ be as above. From what we have proven already, we find that $L$ is stable under all complex conjugations. Assume that $L$ is totally complex; this implies that $K$ is also totally complex. Then, if $n=[K:\mathbb{Q}]$, we have:

$$ 2nw(1)=\sum_{\iota}(f_{\iota}+f_{\bar{\iota}})=\sum_{\iota}(f_{\sigma\iota}+f_{\sigma\bar{\iota}})=2nw(\sigma). $$

So, in this case, the independence is clear.

The assertion that allows us to reduce this to Dirichlet's unit theorem is the following: It suffices to show that, for all $\sigma\in Aut(\mathbb{C})$, $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$.

To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ depends only on $c$ and not on $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have: $$ f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}. $$ Another way to write this is, for all $\iota$: $$ \sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}. $$

So $\sigma$ stabilizes $f$ if and only if it stabilizes $c(f)$, for all complex conjugations $c$. This shows that the stabilizer of $f$ is the absolute Galois group of a CM extension $L/\mathbb{Q}$.

Note that the proof shows that, if the CM sub-field $L$ is totally real, then $f_{\iota}$ is actually independent of $\iota$. In this case, the Hecke character must be the twist of a finite order character by a power of the norm character.

The assertion that allows us to reduce this to Dirichlet's unit theorem is the following: It suffices to show that, for all $\sigma\in Aut(\mathbb{C})$, $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ and $\sigma$.

(EDIT: It was incorrectly stated in the original version that it was enough to require independence from $\iota$ alone).

To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ is independent of $c$ and $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have: $$ f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}. $$ Another way to write this is, for all $\iota$: $$ \sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}. $$

So $\sigma$ stabilizes $f$ if and only if it stabilizes $c(f)$, for all complex conjugations $c$. Let $L$ be the fixed field of the stabilizer of $f$ in $G_{\mathbb{Q}}$. We have shown that $L$ is stabilized by all complex conjugations. To show that $L$ is CM we need to now need to know that all complex conjugations have the same action on $L$. This amounts to showing that $f_{c(\iota)}$ does not depend on the choice of $c$. But, since $f_{\iota}+f_{c(\iota)}$ does not depend on $c$ by hypothesis, this is clear.

(EDIT: The remainder of the proof has been changed to reflect the correction in the criterion above.)

Set $w(\sigma)=f_{\sigma\iota}+f_{\sigma\bar{\iota}}$: this doesn't depend on $\iota$. To finish the proof, we must show that it is independent of $\sigma$ as well. First, let $L$ be as above. From what we have proven already, we find that $L$ is stable under all complex conjugations. Assume that $L$ is totally complex; this implies that $K$ is also totally complex. Then, if $n=[K:\mathbb{Q}]$, we have: $$ 2nw(1)=\sum_{\iota}(f_{\iota}+f_{\bar{\iota}})=\sum_{\iota}(f_{\sigma\iota}+f_{\sigma\bar{\iota}})=2nw(\sigma). $$ So, in this case, the independence is clear.

To finish, it is enough to show the following: If $L$ has one real place, then $L=\mathbb{Q}$. But the hypothesis implies that there is a complex conjugation $c$ that acts trivially on $L$, and so $2f_{\iota}=f_{\iota}+f_{c(\iota)}$ is independent of $\iota$. In this case, the Hecke character must be the twist of a finite order character by a power of the norm character.