Now that I've had the chance to look at Fargues's proof carefully, it seems incomplete for what it claims to prove, but it might answer the question under consideration.

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):

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

Then I claim that there exists a CM sub-field $L\subset K$ such that $f_{\iota}$ only depends on $\iota\vert_L$ (for the purposes of this statement, totally real fields are CM).

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.

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 $0$.

This shows that, for all $(b_v)\in H$,

$$\sum_{v}\frac{f_{\iota(v)}+f_{\overline{\iota}(v)}}{2}b_v=0.$$
Here, if $v$ is complex, $\iota(v)$ and $\overline{\iota}(v)$ are the two complex embeddings inducing $v$. If $v$ is real, they're the same embedding. If one uses the usual basis of $H$ consisting of differences of the basis vectors for the ambient space, one finds that $f_{\iota}+f_{\overline{\iota}}$ is independent of $\iota$. Applying the same argument to the character $\epsilon\mapsto\prod_{\iota}\sigma^{-1}(\iota(\epsilon))^{f_{\iota}}=\prod_\iota\iota(\epsilon)^{f_{\sigma\iota}}$, one sees that $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ as well.

(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.