What construction gives the syntax of an algebraic structure? Let's say I have some algebraic structure $M$ (let's use a monoid for simplicity). I have a specific subset $S \subset M$. The free monoid on $S$ encodes what elements of $M$ I can get by combining elements of $S$.
I'm interested not only on what elements of $M$ I can get, but also on how I can get them. In this case the set of binary trees with elements of $S$ at their leaves.
So, If the Free construction gives me the semantics, what construction gives me the syntax?
 A: In this particular case you could just look at the free magma (non-associative object with binary operation) on $S$. In general contained in any algebraic theory $\mathcal T$ is a subtheory $\mathcal T’$ with the same operations but no axioms, and you could look at the free $\mathcal T’$-object on a subobject.
A: I may disagree on the fact that the free construction gives semantics. Let me try to exlain why. Consider $\mathcal{T}$ the theory of monoids. In my understanding, the free construction gives the syntax. Given a set $X$, note by $\mathcal{T}(X)$ the free object on $X$. It is be the set of expressions you can form using elements of $X$ and the language of monoids, note that it contains also the empty expression. So in the end it sums up to expressions $x_1\cdot x_2\cdots x_n$, hence $\mathcal{T}(X) \simeq \coprod_{n\in \mathbb{N}}X^n$, the empty expression beeing represented by $X^0$, a set with one element. The semantics is the algebra structure : a $\mathcal{T}$-algebra on the set $X$ is the data of a map of sets $a : \mathcal{T}(X)\to X$ that interprets purely syntactic expressions, i.e. elements of $\mathcal{T}(X)$ as elements of $X$.
The data of an $\mathcal{T}$-algebra is equivalent to the data of monoid, since restricting the strucutre map $a$ to $X\times X \subset \mathcal{T}(X)$, gives the multiplication $m:X\times X \to X$, the associativity and unitarity follows form the algebra axioms, and given an associative monoid, we can build the structure map using recursion on the length of the expressions, since expressions of length one are just elements of the monoid, and the empty expression should be sent to the identity of the monoid under the axioms $a$ has to follow.
If I have a monoid, e.g. $(\mathbb{N},\cdot,1)$, I can interpret syntactic expressions in the language of monoids like $2\cdot 3\cdot 7$ (which are in correspondence with the free monoid on $\mathbb{N}$, namely $\mathcal{T}(\mathbb{N})$), as $42$ using the multiplication, and the recursive algorithm.
So in your example, the free monoid on $S$ is just the set of expressions you can form using elements of $S$. Then using the algebra structure on $M$ you can see the image of $\mathcal{T}(S) \subset \mathcal{T}(M)$ under the structure map $f:\mathcal{T}(M) \to M$ as the set of elements in $M$ you can get combining the elements of $S$. Lets call this sub set $\mathcal{I}(S)$ (note that is a monoid since the empty expression is interpreted as the unit of the monoid), so it is the monoid generated by $S$ in $M$.
If you want the set of the expressions that gave you $\mathcal{I}(S)$, then you are looking for the set $f^{-1}(\mathcal{I}(S))$. Note here that since we are considering the theory of monoids, the set $\mathcal{T}(X)$ is the set of unparenthesised expressions since the theory is associative. If you want the set of expressions with parenthesis, or equivalently the set of planar binary trees with leaves decorated by elements of the set $X$, you have to consider the theory $\mathcal{T}'$ which forgets the associativity axiom as Kevin Arlin's answer points out.
Hope this answers somehow your question.
