The category of $T$ algebras on Set is equivalent to the category of monoids Let Set denote the category of sets.
Let $T:$ Set $\to$ Set be the functor that sends a set $X$ to the set of finite words on $X$.
That is, $TX = \{[x_m,..,x_1] : m = 0,1,2,3..., x_i \in X\}$
$T$ can be considered a monad on Set with multiplication given by concatenation, and the unit by $x \to [x]$.

I'm trying to show explicitly that the category of $T$ algebras on Set
  is equivalent to the category of monoids (unital).

To do this we can define a functor $F:$ Mon $\to$ Alg$_T$(Set) by $F(m) = (m,a)$ where $a: Tm \to m$ is s.t $a([f_n,..,f_1]) = f_n...f_1$.
I have shown that this functor is fully faithful.
However I'm not sure I understand if it is essentially surjective.
Letting $(X,a_X) \in$ Alg$_T$(Set). I thought of defining a monoid $m$ as $TX$, where the action is concatenation, and the unit is simply the empty word.
Doing this yields a natural map in Alg$_T$(Set) from $TX$ to $X$, namely $a_X$. Clearly this map is onto because $a_X([x]) = x$ by definition of $(X, a_X)$ being an algebra.
But is this map injective?
 A: You're starting from a $T$-algebra $(X,a_X)$, and you wish to show that it is isomorphic to $F(M)$ for some monoid $M$. 
In particular, since $F(M)$ is of the form $(M,h)$ for $h:TM\to M, [m_1,...,m_n]\mapsto m_1...m_n$, then if there is such an isomorphism $f:X\to M$ and if $x,y\in X$, then $h\circ T(f)([x,y]) = h([f(x),f(y)])=f(x)f(y)$ and $h\circ T(f)([x,y] ) = f\circ a_X([x,y])$. 
Considering that $f$ is a bijection and thus identifying $M$ and $X$ with it, we get a monoid structure on $X$ such that $a_X([x,y]) = xy$. 
Thus now if we want to find $M$, we know what we should do : define multiplication on $X$ by the formula $xy := a_X([x,y])$; and check that this makes $X$ into a monoid such that $F(X) \simeq (X,a_X)$ (actually it will be an equality !)
To prove that $X$ is indeed a monoid with this operation, you will need to use the different properties of the monad and axioms of $T$-algebra. 
Associativity will follow from the associativity axiom of $T$-algebras that says that this should commute :
$\require{AMScd} \begin{CD}
TTX @>{\mu_X}>> TX\\ @V{T(a_X)}VV @VV{a_X}V\\
TX @>>{a_X}> X
\end{CD}$
and the unit will be $a_X([ ])$ and the fact that this is a unit will follow from the associativity axiom for $T$-algebras, together with the unit axiom of $T$-algebras that says that this should commute : 
$\require{AMScd} \begin{CD}
X @>{\eta_X}>> TX\\ @V{id_X}VV @VV{a_X}V\\
X @>>{id_X}> X
\end{CD}$
You then need to prove that $a_X([x_1,...,x_n]) = x_1...x_n$ but this will follow from associativity again. I'll let you check the details. 
