Can a monoid be build from two monoids? This question is written mainly in the context of CWM. 
Let $\mathcal{B}=\left(\mathcal{B}_{0},\square,e,\alpha,\lambda,\varrho,\gamma\right)$
be a symmetric monoidal category. 
Let $\left(c,\mu_{c},\eta_{c}\right)$
and $\left(d,\mu_{d},\eta_{d}\right)$ be monoids. 
Here $\mu_{c}:c\square c\rightarrow c$,
$\eta_{c}:e\rightarrow c$, $\mu_{d}:d\square d\rightarrow d$ and
$\eta_{d}:e\rightarrow d$ suffice certain conditions. 

Is it possible to construct a monoid $\left(c\square d,\mu,\eta\right)$ in a natural
  way? 

I am thinking of $\eta=\left(\eta_{c}\square\eta_{d}\right)\circ\delta$
where $\delta=\lambda_{e}^{-1}=\varrho_{e}^{-1}:e\rightarrow e\square e$
and an arrow $\mu:\left(c\square d\right)\square\left(c\square d\right)\rightarrow c\square d$
of the form $\mu=\left(\mu_{c}\square\mu_{d}\right)\circ\beta$ where
$\beta:\left(c\square d\right)\square\left(c\square d\right)\rightarrow\left(c\square c\right)\square\left(d\square d\right)$
is a composition of instances of $\alpha$, $\alpha^{-1}$ and $\gamma$.
I can imagine that it would be quite a job to give a full proof or counterexample. However, a yes/no answer will be very much appreciated as well.
This case interests me in the sense that (if it is true) with abstract nonsense
you can prove things like: if $A$ and $B$ are rings then $A\otimes B$
can be recognized as a ring. The same for $K$-algebras.
Thank you in advance.
 A: Yes it is a monoid. The monoid axioms can be checked by drawing large enough commutative diagrams. For example, the associativity diagram (I will simply write $cdc$ instead of $c \square (d \square c)$ etc.)
$$\require{AMScd}\begin{CD}
cdcdcd @>cd\mu>> cdcd \\
@V\mu cdVV @VV\mu V \\
cdcd @>\mu>> cd
\end{CD}$$
may be decomposed as follows:
$$\require{AMScd}\begin{CD}
cdcdcd @>>> cdccdd @>cd\mu_c\mu_d>> cdcd \\
@VVV @VVV @VVV \\
ccddcd @>>> cccddd @>c\mu_cd\mu_d>> ccdd \\
@V\mu_c\mu_dcdVV @V\mu_cc\mu_ddVV @VV\mu_c\mu_dV \\
cdcd @>>> ccdd @>\mu_c\mu_d>> cd
\end{CD}$$
The arrows without name are the obvious symmetry isomorphisms (the unique ones which only interchange $c$ and $d$, not $c$ with $c$ or $d$ with $d$).
The square on the upper left commutes because of the Coherence Theorem. The square on the upper right commutes because of the naturality of the symmetry isomorphisms. The same is true for the square on the lower left. The square on the lower right commutes because of the associativity diagrams for $\mu_c$ and $\mu_d$:
$$\require{AMScd}\begin{CD}
ccc @>c\mu_c>> cc @.@.@. ddd @>d\mu_d>> dd \\
@V\mu_ccVV @VV\mu_cV @.@. @V\mu_ddVV @VV\mu_dV \\
cc @>\mu_c>> c @.@.@. dd @>\mu_d>> d
\end{CD}$$
You have already observed that this construction unifies the tensor product of rings, algebras, but also of monoids (here $\otimes=\times$) and of graded algebras (here it is nice to see how the symmetry appears in the definition of the multiplication, namely $(x \otimes y) \cdot (x' \otimes y') := (-1)^{|y||x'|} xx' \otimes yy'$.)
