Let $T$ denote an algebraic theory, and $e$ denote the $T$-algebra freely generated by a singleton set, and write $U$ for the forgetful functor. Now suppose we're given a $T$-algebra $A.$ We might say that an element of $A$ (as opposed to an element of $U(A)$) is just a morphism $f : e \rightarrow A$. Let us write $f \in A$ to mean that $f$ is an element of $A$, and $x \in U(A)$ to mean that $x$ is an element of $U(A)$.
Now lets fix some notation; $e$ is generated by a singleton set, so lets go ahead and call the sole element of that set $k$.
Then every $x \in U(A)$ determines some $f \in A$, namely it is the unique $f : e \rightarrow A$ satisfying $f(k)=x$. Similarly, every $f \in A$ determines some $x \in U(A)$, namely $f(k)$. These processes should be invertible.
This suggests the following. Let $\mathbf{T}$ denote the category of $T$-algebras. It should (essentially) be possible to recover the canonical forgetful functor $\mathbf{T} \rightarrow \mathbf{Set}$ just by knowing which object of $\mathbf{T}$ is the free object on one generator.
Question. Where can I get more information?
The forgetful functor is $\hom(e,-)$. As Qiaochu said, this is essentially the definition of the free algebra on one generator.