No.

Consider an irreducible polynomial $f$ of degree $\geq 6$ with all the nonvanishing coefficients in even degrees. For such a polynomial, for every root $\alpha$, $-\alpha$ is also a root, and thus a Galois conjugate. If $S$ consists of the embeddings that send $\lambda$ to any number of roots together with their negations, then the sum is $0$.

Thus, $2^{ \deg f/2}$ different subsets all lead to the same sum $0$. Some of these are cosets. However, since $\deg f \geq 6$, the subsets of size $\deg f-2$ cannot be cosets as their order doesn't divide $\deg f$, so the map is not injective away from cosets.

No.

Let $a, b \in \mathbb Q(i)$. Let $\alpha_1$ be a root of $x^3 + ax + b$. Let $L$ be the field generated by $i, \alpha_1, \overline{\alpha}_1$. Assume that $a,b$ are sufficiently general that $L$ has Galois group $S_3 \wr \mathbb Z/2$, i.e. $(S_3 \times S_3 ) \rtimes \mathbb Z/2$.

Let $K$ be the subfield generated by $\alpha_1 \overline{\alpha}_1$. Then $K$ is stabilized by $S_2 \wr \mathbb Z/2$, which is a maximal subgroup of index $9$, so $K$ has no proper subfields other than $\mathbb Q$.

Choose $\sigma_1,\sigma_2,\sigma_3$ embeddings which send $\alpha$ to the the three roots $\alpha_1,\alpha_2,\alpha_3$ of $f$ but preserve $\overline{\alpha}_1$ (possible by our assumption on the Galois group.

Then $$\sigma_1( \alpha_1 \overline{\alpha}_1) + \sigma_2( \alpha_1 \overline{\alpha}_1)+ \sigma_3( \alpha_1 \overline{\alpha}_1)=\alpha_1 \overline{\alpha}_1 + \alpha_2 \overline{\alpha}_1 + \alpha_3 \overline{\alpha}_1 = (\alpha_1 + \alpha_2 +\alpha_3) \overline{\alpha}_1 = 0 \overline{\alpha}_1=0$$

so $x\mapsto \sigma_1(x)+\sigma_2(x) + \sigma_3(x)$ is not injective.