(I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!)
The situation is even better than that! Suppose we are given an $E_1$-algebra $A$ of a presentable symmetric monoidal $\infty$-category $\mathcal{C}$.
Call an $E_n$-monoidal structures on the $\infty$-category $\mathbf{Mod}(A)$ of left $A$-modules allowable if $A$ is the unit and the right action of $\mathcal{C}$ on $\mathbf{Mod}(A)$ is compatible with the $E_n$ monoidal structure, so that $\mathbf{Mod}(A)$ is an $E_n$-$\mathcal{C}$-algebra. Then the space of allowable $E_n$-monoidal structures is equivalent to the space of $E_{n+1}$-algebra structures on $A$ itself, compatible with the extant $E_1$ structure on $A$. (This is even true when $n=0$, if one takes an $E_0$-monoidal category to mean a category with a distinguished object.) The object $A$, regarded as the unit $A$-module, admits an $E_n$-algebra structure that is suitably compatible with the $E_1$ structure an $A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.]
Let's sketch a proof of this claim in the case Peter mentions. Suppose $A$ is a monoid in a presentable symmetric monoidal category $(\mathbf{C},\otimes)$. Suppose $\mathbf{Mod}(A)$ admits a monoidal structure (not even a priori symmetric!) in which $A$, regarded as a left $A$-module, is the unit. I claim that $A$ is a commutative monoid. Consider the monoid object $\mathrm{End}(A)$ of endomorphisms of $A$ as a left $A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give $\mathrm{End}(A)$ the structure of a commutative monoid object. The multiplication on $A$ yields an isomorphism of monoids $A\simeq\mathrm{End}(A)$.
In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If $X$ admits magma structures $\circ$ and $\star$ with the same unit (Below, Tom Leinster points out that I only have to assume that each has a unit, and it will follow that the units are the same. He's right, of course.) with the property that
$$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$
for any $a,b,c,d\in X$, then (1) the magma structures $\circ$ and $\star$ coincide; (2) the product $\circ$ is associative; and (3) the product $\circ$ is commutative. That is, a unital magma in unital magmas is a commutative monoid.
First, having seen the edited version of Martin's question, let's quickly dispose of the construction of the free symmetric monoidal category generated by a category $C$. Objects are tuples $(x_1, \ldots, x_n)$ of objects of $C$. Morphisms are labeled permutations, where permutations are conveniently visualized as string diagrams, each string being labeled by a morphism in $C$. To compose such labeled diagrams, just compose the string diagrams, composing the labels of strings in $C$ along the way.
The free symmetric monoidal category $Sym(M)$ on a monoidal category $M$ is formed from the free symmetric monoidal category $S (U M)$ on the underlying category $U M$ by adjoining isomorphisms $\phi_{x_1, \ldots, x_n}: (x_1, \ldots, x_n) \to x_1 \otimes \ldots \otimes x_n$, where the $x_i$ are objects of $M$, the expression $(x_1, \ldots, x_n)$ is the formal monoidal product in $S(UM)$, and $x_1 \otimes \ldots \otimes x_n$ is the tensor product in $M$.
More precisely, define the objects of $Sym(M)$ to be tuples $(x_1, \ldots, x_n)$ of objects of $M$. Define morphisms $(x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ to be equivalence classes of pairs $(p, f)$ where $p$ is a permutation on $n$ elements, and $f$ is an $M$-labeled planar forest of $m$ rooted trees. (You should think here of the free multicategory generated by a category.) Formally, a planar forest can be described as a functor $[n]^{op} \to \Delta$ where $n$ is the category $1 \leq \ldots \leq n$ and $\Delta$ is the category of finite (possibly empty) ordinals. Under the obvious way of drawing such forests, the ordered list of leaves of the forest is labeled by $(x_{p(1)}, \ldots, x_{p(n)})$, and the ordered list of roots by $(y_1, \ldots, y_m)$. Edges are labeled by objects of $M$, and each internal node with $k$ inputs labeled (in order) by $m_1, \ldots, m_k$ and output $m$ is labeled by a morphism $f: m_1 \otimes \ldots \otimes m_k \to m$.
There is an evident way, using the monoidal structure of $M$, of evaluating such a labeled forest $f$ as a morphism $ev(f): x_{p(1)} \otimes \ldots \otimes x_{p(n)} \to y_1 \otimes \ldots \otimes y_m$ in $M$. We consider two arrows $(p, f)$ and $(p', f')$ to be equivalent if $p = p'$ as permutations and $ev(f) = ev(f')$.
Now we define composition of pairs. The main idea is to rewrite the composition of a forest followed by a permutation,
$$(x_1, \ldots, x_n) \stackrel{(1, f)}{\to} (y_1, \ldots, y_m) \stackrel{(q, 1)}{\to} (y_{q(1)}, \ldots, y_{q(m)}),$$
into a form $(p, f')$ forced by the naturality requirement of the symmetry isomorphism. Namely, if $\bar{x}_i$ is the tuple of leaves of the tree whose root is $y_i$, then we have a block permutation $bl(q)$ taking $(\bar{x}_1, \ldots, \bar{x}_m)$ to $(\bar{x}_{q(1)}, \ldots, \bar{x}_{q(m)})$. The permutation $q$ can also be applied to the $m$ trees of the forest by reordering the trees, yielding a new forest $\mathrm{perm}_q(f)$, and the composition $(q, 1) \circ (1, f)$ is defined to be $(bl(q), \mathrm{perm}_q(f))$. Then, if we have $(p, f): (x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ and $(q, g): (y_1, \ldots, y_m) \to (z_1, \ldots, z_p)$, we define their composite to be
$$(bl(q) \circ p, g \circ \mathrm{perm}_q(f))$$
where the first component is by composing permutations and the second is by the usual way of composing forests by plugging in roots of one planar forest of trees into leaves of another.
Remaining details that all this works will be left to the reader. I will remark that the key rewriting procedure above is an instance of a kind of distributive law; there are some more details to this effect in some notes on my nLab web; see here.
Best Answer
I think you are interpreting it as "(free monoidal category) on a monoid", while it's really "free (monoidal category on a monoid)", or rather "free (monoidal category with a monoid)".
Now to construct it you have have to find which morphisms should exists as a consequences of the axioms, i.e. which morphisms you are sure to see whenever you have a monoid in a monoidal category. Clearly, if $X$ is an object in a monoidal category $C$, the only morphisms between $X^{\otimes n}$ and $X^{\otimes m}$ that exists whatever $(C,X)$ are, are the identity when $m=n$. So the free monoidal category is just $\mathbb{N}$ whith $\hom(m,n)=${id} if $m=n$, and is empty otherwise.
But if $X$ is a monoid, you get by definition a morphism $X \otimes X \rightarrow X$ and a morphism $I \rightarrow X$ where $I$ is the identity object. These morphisms clearly extends to morphisms $\delta_i:X^{\otimes n+1}\rightarrow X^{\otimes n}$ and $\sigma_i: X^{\otimes n}\rightarrow X^{\otimes n+1}$ just by tensoring them with identity morphisms, and then you can compose them to get maps $X^{\otimes m}\rightarrow X^{\otimes n}$. Finally, it's easy to see that the commutation relation between these morphisms imposed by the axioms of a monoid are precisely those satisfied by the simplicial maps in $\Delta$.