A prime splits completely in $L$ over $K$ (an extension of number fields) if and only if it splits completely in the Galois closure of $L$ over $K$. Thus to answer the question we may assume that $L$ is Galois over $K$.
Ben Linowitz's argument then holds in generality: suppose that all $\wp$ congruent to $1$ modulo some conductor $\mathfrak m$ split in $L$. Then by the Lemma in Ben's answer, $L$ is contained in the ray class field of conductor $\mathfrak m$ over $K$, and hence is abelian.
(As far as I can tell, this is not at all obvious without class field theory, and in fact, a big part of the development of class field theory involved the realization that class fields --- which were defined in terms of splitting conditiosn described by congruences --- were the same things as abelian extensions. In some sense, the equivalence of these two conditions is the essence of class field theory.)
[EDIT:] This edit is in response to Buzzard's comments on the original question, and also his answer and subsequent comments.
Suppose that $L$ over $K$ is Galois (as we may) and that for some non-empty subset $S$ in some ray class group $Cl_{\mathfrak m}$ we know that all (but finitely many) primes lying in $S$ mod $\mathfrak m$ split in $L.$
If we furthermore assume if and only if in the preceding statement, then Buzzard's answer shows that $S$ must contain the trivial class, and hence $L$ is an abelian extension contained in the ray class field of conductor $\mathfrak m$; class field theory then takes over to show that $S$ is in fact a subgroup.
But what if we don't assume if and only if (i.e. we allow that other primes besides those lying in $S$ split)? Can we still argue that $L$ is abelian over $K$?
Let $L'$ be the compositum of $L$ and the ray class field of conductor $\mathfrak m$ over $K$, let $G = Gal(L/K)$, and let $G' = Gal(L'/K)$.
Then $G' \hookrightarrow G \times Cl\_{\mathfrak m}$ (via the Galois action on $L'$ and the ray class field resp.); let $p$ and $q$ be the first and second projections (note that they are both surjective).
Our assumption translates into the statement that $p (q^{-1}(S)) = \{1\}$, i.e. $q^{-1}(S) \subset 1 \times Cl\_{\mathfrak m}.$
Now choose $s \in S$, and suppose that $(g,1) \in G'$. The previous paragraph together with the surjectivity of $q$ shows that also $(1,s) \in G'$. Then $(g,1) (1,s) = (g,s) \in G',$ since $G'$ is a subgroup of the product. But $(g,s)$ lies in $q^{-1}(S)$, hence $g = 1$. In other words, if the second coordinate of an element of $G'$ is trivial, so is the first. Thus in fact $L'$ equals the ray class field of conductor $\mathfrak m,$ i.e. $L$ is contained in the latter field. This is what had to be shown.