Let $X$ be any set. A monoid $M$ is called universal over $X$ iff $X \subseteq M$ and for every other monoid $N$ and function $\varphi : X \to N$ there exists a unique extension $\varphi : M \to N$ of $\varphi : X \to N$ which is a homomorphism from $M$ to $N$.
Now I want to show that $$ M \mbox{ is universel iff $M = \langle X \rangle$ and no nontrivial relations hold among its elements } $$ or to be more precise, iff $M = \langle X \rangle$ and if $x_i, y_j \in X$ with $i \in \{1,\ldots, n\}, j \in \{1,\ldots, m\}$ the equation $$ x_1 x_2 \cdots x_n = y_1 y_2 \cdots y_m $$ implies $n = m$ and $x_i = y_i$ for $i = 1,\ldots, n$.
Now the only way that comes to my mind to prove this is to give an explit reference object, like $X^{\ast}$ (the set of finite sequences), and show that if such an relation holds in $M$, then two distinct sequences in $X^{\ast}$ would be equal (by computing with the unique extensional homomorphism).
But is there any more intrinsic way to show this, without using a concrete object (which in some way already is the universal object), i.e. using the definition of universality in a more fundamental way?
To give an example what I have in mind: That $M$ is determined up to isomorphism could also be proven by giving an explicit isomorphism onto $X^{\ast}$, but it could also be proven by applying the definition without any reference to an external object, i.e. if $M$ and $N$ are universal over $X$ apply it to $\mbox{id} : X \to X \subseteq N$ to get a unique extension $\varphi : M \to N$ and also $\mbox{id} : X \to X \subseteq M$ to get a unique extension $\psi : N \to M$, then $\varphi\circ\psi$ is an extension of $\mbox{id} : X \to X$ when we apply the definition to $M$ itself, and also as $\mbox{id} : M \to M$ is such an extension uniqueness gives $\varphi\circ\psi = \mbox{id}$, and similar we find $\psi\circ\varphi = \mbox{id}$, so $M \cong N$.
In some sense you can't do much better than use $X^*$. In fact, that no non-trivial relation holds in a free monoid is equivalent to the fact that for any relation, there is at least one monoid generated by $X$ in which this precise relation doesn't hold.
Indeed, in any category $C$, if $P$ is a property that is transfered by morphisms (ie if an object $A$ has the property $P$ and there is a morphism $A\to B$ then $B$ has $P$), then an initial object of $C$ has the property $P$ iff all objects have it.
And here $M$ is an initial object in the category of monoids $N$ equipped with a function $X\to N$, morphism being monoid morphisms which are compatible with the functions from $X$.
So since the fact that some relation involving elements of $X$ is transfered by morphisms in this category, then for any relation, it holds in $M$ iff it holds in all monoids generated by $X$ ; in other words, in doesn't hold in $M$ iff it doesn't hold in some monoid generated by $X$.
So the fact that we can build an explicit monoid $X^*$ in which no non-trivial relation holds implies than no non-trivial relation holds in $M$ (and that actually shows that $M\simeq X^*$). But even if we hadn't had the idea of this example that works for all relations, we could have constructed one example for each relation (no sure that's much more simple, though).