Coherence in closed monoidal categories Let $(M, \otimes, I)$ be a left-closed (non-symmetric) monoidal category with left-internal hom  $\underline{\operatorname{hom}}(-,-)$.
Denote by $\sigma_{A,B,C}: M(A\otimes B, C) \xrightarrow{\sim} M(A,\underline{\operatorname{hom}}(B,C))$ the natural isomorphism obtained from the closed monoidal structure.
Consider the evaluation morphism $\operatorname{ev}_{A,B}: \underline{\operatorname{hom}}(A,B) \otimes A \rightarrow B$.
Define for $A,B \in M$ the following morphism $f_{A,B}:= \sigma_{A \otimes \underline{\operatorname{hom}} (I,B),I,A\otimes B}(\operatorname{id}_A \otimes\operatorname{ev}_{I,B})$.
For $A \in M$ let $g_A: \underline{\operatorname{hom}}(I,A) \xrightarrow{\sim} A$ be the isomorphism obtained from the Yoneda lemma.
I am struggling to show that the following equality of morphisms holds for any two objects $A,B \in M$: $$g_{A \otimes B} \circ f_{A,B}=\operatorname{id}_A \otimes g_B.$$
Does this equality even hold? Any ideas on how to prove it?
 A: Drawing all the commutative diagrams and writing all the subscripts is too troublesome, so I will briefly sketch the proof in words instead.
Hopefully you will be able to reconstruct the details yourself.
Consider the right unitor $\rho : X \otimes I \to X$.
It corresponds under the tensor–hom adjunction to $[I, \rho] \circ \eta : X \to [I, X]$, where $\eta : X \to [I, X \otimes I]$ is the unit of the adjunction.
You know that $X \to [I, X]$ is invertible: indeed, its inverse is the isomorphism $g : [I, X] \to X$ you ask about.
(Well, I believe it is.
Your definition is dependent on the choice of natural isomorphism $X \otimes I \cong I$ but I presume you have made the only truly natural choice, namely $\rho$.)
So we find that $g = \epsilon \circ \rho^{-1}$, where $\epsilon : [I, X] \otimes I \to X$ is the counit (i.e. evaluation) and $\rho^{-1} : [I, X] \to [I, X] \otimes I$ is the inverse unitor.
The morphism $f : A \otimes [I, B] \to [I, A \otimes B]$ you define is given by $[I, (A \otimes \epsilon) \circ \alpha] \circ \eta$, where $\alpha : (A \otimes [I, B]) \otimes I \to A \otimes ([I, B] \otimes I)$ is the associator.
Thus, your question becomes: is
$$
A \otimes [I, B] 
\xrightarrow{A \otimes \rho^{-1}}
A \otimes ([I, B] \otimes I)
\xrightarrow{A \otimes \epsilon}
A \otimes B
$$
equal to the following composite?
\begin{multline}
A \otimes [I, B] 
\xrightarrow{\eta} 
[I, (A \otimes [I, B]) \otimes I] 
\xrightarrow{[I, \alpha]} [
I, A \otimes ([I, B] \otimes I)]
\xrightarrow{[I, A \otimes \epsilon]}
[I, A \otimes B] \\
\xrightarrow{\rho^{-1}}
[I, A \otimes B] \otimes I
\xrightarrow{\epsilon}
A \otimes B
\end{multline}
The answer is yes, of course.
Since all of the morphisms involved are invertible, it suffices to show that the composite
\begin{multline}
A \otimes [I, B] 
\xrightarrow{\eta} 
[I, (A \otimes [I, B]) \otimes I] 
\xrightarrow{[I, \alpha]} [
I, A \otimes ([I, B] \otimes I)]
\xrightarrow{[I, A \otimes \epsilon]}
[I, A \otimes B] \\
\xrightarrow{\rho^{-1}}
[I, A \otimes B] \otimes I
\xrightarrow{\epsilon}
A \otimes B
\xrightarrow{A \otimes \epsilon^{-1}}
A \otimes ([I, B] \otimes I)
\xrightarrow{A \otimes \rho}
A \otimes [I, B]
\end{multline}
is the identity.
Using naturality to shuffle $\rho$ around, it eventually boils down to two identities:
$$
(A \otimes [I, B]) \otimes I
\xrightarrow{\alpha}
A \otimes ([I, B] \otimes I)
\xrightarrow{A \otimes \rho}
A \otimes [I, B]
$$
is equal to $\rho : (A \otimes [I, B]) \otimes I \to A \otimes [I, B]$, and
$$
(A \otimes [I, B]) \otimes I
\xrightarrow{\eta \otimes I}
[I, (A \otimes [I, B]) \otimes I] \otimes I
\xrightarrow{\epsilon}
(A \otimes [I, B]) \otimes I
$$
is the identity.
A: I am not finished yet and will later revisit my answer, but for now I will share, what I got so far:
The given equation is equivalent to $f_{A,B}\circ(\operatorname{id}_A\otimes g_B^{-1})=f_{A,B}\circ(\operatorname{id}_A\otimes g_B)^{-1}=g_{A\otimes B}^{-1}$.
Let $\rho_A\colon A\otimes I\rightarrow A$. $g_A^{-1}$ and $\operatorname{ev}_{A,B}$ correspond to the identity under the $-\otimes X\dashv\underline{\operatorname{hom}}(X,-)$ adjunction:
\begin{equation}
\Phi_A\colon
\underline{\operatorname{hom}}(A,A)\xrightarrow{\underline{\operatorname{hom}}(\rho_A^{-1},A)}\underline{\operatorname{hom}}(A\otimes I,A)\xrightarrow{\sigma_{A,I,A}}\underline{\operatorname{hom}}(A,\underline{\operatorname{hom}}(I,A)),\operatorname{id}_A\mapsto g_A^{-1}
\end{equation}
\begin{equation}
\Psi_{A,B}\colon
\underline{\operatorname{hom}}(\underline{\operatorname{hom}}(A,B),\underline{\operatorname{hom}}(A,B))
\xrightarrow{\sigma_{\underline{\operatorname{hom}}(A,B),A,B}^{-1}}\underline{\operatorname{hom}}(\underline{\operatorname{hom}}(A,B)\otimes A,B),
\operatorname{id}_{\underline{\operatorname{hom}}(A,B)}\mapsto\operatorname{ev}_{A,B},
\end{equation}
I assume we have $\Phi_A=\underline{\operatorname{hom}}(A,g_A^{-1})$ and therefore $g_A^{-1}=\underline{\operatorname{hom}}(A,g_A^{-1})(\operatorname{id}_A)$ as well as $\underline{\operatorname{hom}}(\operatorname{ev}_{A,B},B)(\operatorname{ev}_{A,B})=\operatorname{id}_B$. I also assume we have:
\begin{equation}
\operatorname{id}_A
=\operatorname{ev}_{I,A}\circ\rho_A^{-1}\circ g_A\colon
A\rightarrow\underline{\operatorname{hom}}(I,A)
\rightarrow\underline{\operatorname{hom}}(I,A)\otimes I
\rightarrow A.
\end{equation}
Also, given a map $f\colon A\otimes B\rightarrow C$, we have $\operatorname{ev}_{B,C}\circ(\sigma_{A,B,C}(f)\otimes\operatorname{id}_B)=f$. We now have:
\begin{align*}
f_{A,B}\circ(\operatorname{id}_A\otimes g_B^{-1})
=\sigma_{A\otimes\underline{\operatorname{hom}}(I,B),I,A\otimes B}\big(\operatorname{id}_A\otimes\Psi_{I,B}(\operatorname{id}_{\underline{\operatorname{hom}}(I,B)})\big)
\circ\Big(\operatorname{id}_A\otimes\Phi_B(\operatorname{id}_B)\Big)
=\ldots \\
\end{align*}
