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 $\iota$.
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 $L/\mathbb{Q}$.
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.
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.