Algebraic Functors and Left Adjoints

You're correct, the left adjoint isn't the group ring construction. The underlying additive group functor $\text{Ring} \to \text{Group}$ splits as a composite

$$\text{Ring} \to \text{Ab} \to \text{Grp}$$

so its left adjoint splits as a composite going the other way. The left adjoint of $\text{Ab} \to \text{Grp}$ is abelianization, and the left adjoint of $\text{Ring} \to \text{Ab}$ is the tensor algebra functor. Their composite sends a group $G$ to the tensor algebra

$$T(A) = \bigoplus_{n \ge 0} A^{\otimes n}$$

of the abelianization $A = G/[G, G]$.

Similarly the left adjoint to the underlying additive group functor from commutative rings to abelian groups is given by taking the symmetric algebra

$$S(A) = \bigoplus_{n \ge 0} A^{\otimes n} / S_n.$$

The free ring (with unit, noncommutative) on an abelian group $A$ is the tensor algebra $\sqcup A^i$, with multiplication given by concatenation. The left adjoint to the forgetful functor from rings to groups just composes this with abelianization. Left adjoints to algebraic functors are always free constructions: in this case, one simply adds in whatever products are needed. This is indeed not the group ring. One should specify that $F$ sends the multiplication in the syntax of groups to the addition in the syntax of rings, so that a connection to group rings, where the group structure is Incorporated via multiplication, wouldn't make sense.

Regarding the group ring, it cannot be the left adjoint of an algebraic functor, because it doesn't preserve retracts of finitely generated free objects. Indeed, $\mathbb Z[X,X^{-1}]$, the group ring of $\mathbb{Z}$, is not even a subring of any free ring, since free rings have finite groups of units. The group ring is, however, the left adjoint of an accessible functor, one living in the doctrine of locally (finitely) presentable categories, as it does preserve finitely presentable objects.

An excellent reference on algebraic categories is Algebraic Theories, by Adamek, Rosicky, and Vitale.