I don't see why the restriction to galois extensions is necessary. Consider, for example, the non-galois cubic field $K = \mathbb Q(\sqrt[3]{2})$. Then no prime congruent to 2 mod 3 splits completely in $K$. Indeed, if $p\equiv 2\pmod 3$, then 2 has a unique cube root mod p, and so the polynomial $x^3 - 2$ factors mod p into a linear term times a quadratic. Now the ring of integers of $K$ turns out to be $\mathbb Z[\sqrt[3]{2}]$, and therefore it follows that p factors in $K$ as a product of two primes of residue degrees 1 and 2.
More generally if $K/\mathbb Q$ is a cubic extension of discriminant $d$, and if $p$ is unramified in $K$ and factors into $g$ primes there, then a formula of Stickelberger tells us that $(\frac{d}{p})=(-1)^{3-g}$. So if $(\frac{d}{p})=-1$ then $p$ factors in $K$ as a product of two primes. Quadratic reciprocity allows us to rewrite the condition $(\frac{d}{p})=-1$ in terms of congruences mod $4d$ (for odd $p$, at least, but that's good enough).
Of course if we know that $K$ is abelian, so that $K \subseteq \mathbb Q(\zeta_n)$ for some $n$, then we can use the cyclotomic decomposition laws to see that the primes that split completely in $K$ are given by congruence conditions mod $n$. (One can be more precise here: a prime splits completely in $K$ iff it belongs to one of the congruence classes in $\text{Gal}(\mathbb Q(\zeta_n)/K) \subset \text{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong (\mathbb Z/n\mathbb Z)^\times$.) Consequently, the leftover congruence classes will not contain any primes that split completely.
