Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.
> K<s>:=QuadraticField(-23);
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);
> jrel:=PowerRelation(jinv,3 : Al:="LLL");
> Kj<j>:=ext<K|jrel>;
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23
> c4, c6 := Explode(cInvariants(E)); // random twist with this j
> f:=Polynomial([-c6/864,-c4/48,0,1]);
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic
> R:=Roots(poly);
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];
> GaloisGroup(f); /* not immediate to compute */
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);
false
This group has $A_4$ and $Z_2^4$ as normal subgroups, but I don't know it's name if any.
PS. 5-torsion is too long to compute most often.
(modified)
Historically, Complex Multiplication precedes Class Field Theory and many of the main theorems of CM for elliptic curves were proved directly. See Algebren (3 volumes) by Weber or Cox's book for an exposition.
Please also read Birch's article on the beginnings of Heegner points where he points this out explicitly (page three, paragraph beginning "Complex multiplication ...).
But not all. so the answer is no (unlike what I had mistakenly presumed at first and the comments below alerted me).
The actual history is quite complicated; see Schappacher.
Best Answer
Just as a minor warning: even if the conductor is $1$, there might be nontrivial roots of unity in the class field: take $K = {\mathbb q}(\sqrt{-5}\,)$ and ${\mathfrak c} = (1)$; then the ray class field is the Hilbert class field $K(\sqrt{-1})$, which contains the 4th roots of unity. The roots of unity in the Hilbert class field (i.e. for conductor $1$) lie in the genus class field and can be computed easily.
Any additional roots of unity must come from ramified extensions; a necessary condition for the $p$-th roots of unity to lie in the ray class field must be that the ry class number, which is easily computed, be divisible by $p-1$ (or $(p-1)/2$ if the genus class field contains the quadratic subfield of the $p$-th roots of unity).