(1) There is a famous Isbell's example (see CFWM, p.164): take $\mathbf{Set}_0$, the skeleton of the category of sets, then for every infinite set $X$ the associator $\alpha_{X,X,X}\colon(X\times X)\times X\to X\times(X\times X)$ cannot be equal to the identity $id_X\colon X\to X$.
(2) A formal diagram is a diagram, which is constructed without using the fact that formally different objects in vertices of a diagram are equal. In the previous example, there was the diagram:
$$
(X\times X)\times X\xrightarrow{\alpha_{X,X,X}}X\times(X\times X)\xrightarrow{id_X} X,
$$
which is constructed using the fact that $X\times(X\times X)=X$, which is an equality between formally different objects.
(3) Free monoidal categories are directly connected with the coherence issues. The coherence theorem for monoidal categories is equivalent to the statement that the free monoidal category generated by an arbitrary set is a preorder. Mac Lane uses another strategy of the proof: he proves the statement for the free monoidal category, generated by the one-elemented set (the set of objects of such free monoidal category is the free magma, generated by the two-elemented set {variable,unit}), and then applies this to the new category $It(B)$, what allows him to prove the whole statement. The proof seems long because it is quite detailed.
I don't know if this is appropriate, but for the proof using free categories generated by the set you can see postings at my categorical weblog (in russian): [1] (construction of the free magma), [2] (for semigroupoidal categories), [3] and [4] (for monoidal categories).
The proof should go as follows.
In the strict monoidal category, we have that the corresponding compositions of $\alpha'$, $\alpha^{\prime -1}$, $l'$, $l^{\prime-1}$, $r'$, $r^{\prime -1}$ are all the identity.
Let $P_{Fi}$ be the corresponding parenthesization of $FX_1,\ldots,FX_n$ and $1'$s to $P_i$.
Monoidality of the equivalence gives natural isomorphisms $\alpha_i : FP_i \to P_{Fi}$ such that
$$
\require{AMScd}
\begin{CD}
FP_1 @>F f>> FP_2\\
@V\alpha_1 VV @V\alpha_2 VV \\
P_{F1} @> \mathrm{id} >> P_{F2}.
\end{CD}
$$
The same diagram commutes for $g$, so $Ff=Fg$.
Then by faithfulness of $F$, $f=g$.
Constructing $\alpha_i$
It sounds like you're having trouble with constructing the $\alpha_i$, so I'll expand on this a bit.
This goes inductively, so we just need to deal with the outermost layer of $P_1$ or $P_2$. We'll just look at $P_1$.
If $P_1 = X\otimes Y$, then $FP_1 = F(X\otimes Y)$ and $P_{F1} = X_F \otimes Y_F$, where the $F$ subscript refers to the appropriate product of $F$ applied to the atoms. Then we have the monoidal structure isomorphism $J : F(X\otimes Y) \to F(X)\otimes F(Y)$, and then we inductively construct $\alpha_X : F(X)\to X_F$ and $\alpha_Y : F(Y)\to Y_F$. The composite $(\alpha_X \otimes \alpha_Y) \circ J$ gives the desired isomorphism $FP_1\to P_{F1}$.
On the other hand, if $P_1 = I\otimes X$, then $P_{F1} = I'\otimes X_F$.
This time, if $\iota : F(I)\to I'$ is the natural isomorphism, we take the composite
$$F(I\otimes X)\overset{J}{\to} F(I)\otimes FX \overset{\iota\otimes \alpha_X}{\to} I'\otimes X_F= P_{F1}$$
For $P_1 = X\otimes I$ we do the same thing but modified for symmetry.
Hopefully this helps. I can expand on this if needed.
Edit, an expansion on why the diagram commutes.
This is also proved inductively.
Suppose the outermost function in $f$ is an associator, which I'll denote by $a$ and $a'$ in the two categories in the equivalence in $C_s$, $a'=\mathrm{id}$, since I used $\alpha$ for the natural isomorphisms.
Then $f=a\circ f_0$ and $f'=a'\circ f_0'$, with $f:P_1\to P_2$ and
$P_2 : X\otimes (Y \otimes Z$ for some products $X$, $Y$, and $Z$,
so $f_0 : P_1\to (X\otimes Y)\otimes Z$.
Then we have the commutative diagram
$$
\newcommand\id{\mathrm{id}}
\begin{CD}
FP_1 @>f_0>> F((X\otimes Y) \otimes Z) @>Fa>> F(X \otimes (Y\otimes Z)) \\
@VVV @VJVV @VJVV \\
@. F(X\otimes Y) \otimes FZ @. FX\otimes F(Y\otimes Z) \\
@V\alpha_1 VV @VJ\otimes \id VV @V\id \otimes J VV \\
@. (FX\otimes FY)\otimes FZ @>a'>> FX\otimes (FY\otimes FZ)\\
@VVV @V(\alpha_X\otimes \alpha_Y)\otimes \alpha_Z VV @V\alpha_X \otimes (\alpha_Y\otimes \alpha_Z)VV \\
P_{F1} @>f'_0>> (X_F\otimes Y_F)\otimes Z_F @>a'>> X_F \otimes (Y_F\otimes Z_F) \\
\end{CD}
$$
The left rectangle commutes by the inductive hypothesis applied to $f_0$, since the middle vertical composite is the definition of the $\alpha$ map for $(X\otimes Y)\otimes Z$. The top right rectangle commutes by the compatibility condition for $J$. The bottom right square commutes by naturality of the associator.
By symmetry, the same argument applies to $a^{-1}$.
Now we need to do the same thing for the left and right units and their inverses. By symmetry, it suffices to prove commutativity when the outermost map in $f$ is $l$.
Then $f = l\circ f_0$, with $f_0 : P_1\to I\otimes P_2$. This time, we get the diagram
$$
\begin{CD}
FP_1 @>f_0>> F(I\otimes P_2) @>Fl>> FP_2 \\
@VVV @VJVV @V\id VV \\
@. FI\otimes FP_2 @. FP_2 \\
@V\alpha_1 VV @V \iota \otimes \id VV @V\id VV \\
@. I'\otimes FP_2 @>l'>> FP_2 \\
@VVV @V\id \otimes \alpha_2 VV @V\alpha_2VV \\
P_{F1} @>f'_0>> I'\otimes P_{F2} @>l'>> P_{F2} \\
\end{CD}
$$
Again, the left rectangle commutes by the inductive hypothesis, the top right rectangle is the coherence condition for $\iota$, and the bottom right square is naturality of $l'$.
This completes the proof.
Best Answer
Technically this is only true for free diagrams, i.e., diagrams which take into account only the associator, right and left unitors, and identity morphisms, but not any other particular isomorphism (or equality between objects) that could be present in your particular category (I'll give a counterexample below).
The actual theorem is then that in the free tensor category generated by a set, every diagram commutes.
This is easy to see as a corollary of the strict tensor equivalence result: the free tensor category $FX$ is tensor equivalent to a strict one $F_sX$, which is discrete (composed only of identity transformations). Therefore:
Now let us see a tensor category with a noncommutative diagram, an example due to Isbell:
Consider the skeleton category of the category Set (which is the full subcategory of cardinal numbers), made a tensor category with the cartesian product $\times$ as tensor product. Let $N$ denote the object with denumerable cardinal; then $N=N\times N$ and both projections $p_1,p_2:N\times N\rightarrow N$ are epi morphisms $N\rightarrow N$. We show that the associator isomorphism $a:(N\times N)\times N=N\times(N\times N)$ is not the identity. Due to the naturalness of $a$, if it was the identity, for any three $f,g,h:N\rightarrow N$ we would have $f\times(g\times h)=(f\times g)\times h$. But then $$fp_1=p_1(f\times(g\times h))=p_1((f\times g)\times h)=(f\times g)p_1:N\rightarrow N$$ implies that $f=f\times g$ for all $f,g$ (because $p_1$ is epi). Similarly $p_2$ being epi means that $g=f\times g=f$, which is a contradiction. So $a_{N,N,N}$ is not the identity. Thus the diagram
is not commutative. The problem here is that $N=N\times N=(N\times N)\times N$ is an identity which is not present in a free tensor category.