First of all, I agree with your update, and I find that that proof is much clearer conceptually ("they're categorical properties so they're preserved under equivalences").
But I also understand why you want "closure" on that first proof; note that the claim isn't so much "the isomorphisms $P\otimes Q = R$ and $Q\otimes P= S$ are such that blabla", but "the isomorphisms can be chosen so that the square commutes" (this is enough for the argument to go through)
I don't have a very easy solution for that but one way to prove that they can be chosen so as to be compatible is as follows :
Let $F$ denotes $P\otimes_S-$ and $G=Q\otimes_R-$. Then by definition $F,G$ form an equivalence between $S-Mod$ and $R-Mod$. Now a standard theorem of category theory states essentially the following : an equivalence can always be made into an adjoint equivalence.
What this means is that we can choose the natural isomorphisms $\eta : id \to GF$ and $\epsilon : FG\to id$ so as to define an adjunction $F\dashv G$.
This implies in particular that they satisfy the triangle identities : $\epsilon F\circ F\eta = id_F$.
Now apply this to $S$ to get the following commutative diagram :
$\require{AMScd}\begin{CD} FGF(S) @<F(\eta_S)<< F(S) \\
@V\epsilon_{F(S)}VV @VVV \\
F(S) @>>> P\end{CD}$
where the two maps $F(S)\to P$ are the same fixed isomorphism (the standard one).
Now since $F(\eta_S)$ is actually an isomorphism, we can reverse it and get :
$\require{AMScd}\begin{CD} FGF(S) @>F(\eta_S^{-1})>> F(S) \\
@V\epsilon_{F(S)}VV @VVV \\
F(S) @>>> P\end{CD}$
And now if you unravel what this is :
$\require{AMScd}\begin{CD} P\otimes_S Q\otimes_R P\otimes_S S @>P\otimes_S\cong \otimes_SS>> P\otimes_S S \\
@V\cong\otimes_R P\otimes_S S VV @VVV \\
R\otimes_RP\otimes_SS @>>> P\end{CD}$
Now of course (this is an easy exercise) you can remove the $\otimes_SS$ from the story and get
$\require{AMScd}\begin{CD} P\otimes_S Q\otimes_R P @>P\otimes_S\cong >> P\otimes_SS \\
@V\cong\otimes_R PVV @VVV \\
R\otimes_RP @>>> P\end{CD}$
which is what you wanted.
So the isomorphisms can be chosen for the diagram to commute, so with these chosen isomorphisms, the argument goes through.
Best Answer
The equivalent proposition is:
Proposition. $R$ and $S$ are Morita equivalent if and only if there is a progenerator $P$ of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$.
While searching for the answer of this question, I found some texts in the internet which stated the above proposition putting $\text{End}(P) \simeq S$ instead of $\text{End}(P) \simeq S^{\text{op}}$, and it is wrong! I thought it was important to share this detail, because it may confuse people, as it confused me. Now let's see the explanation:
The fundamental motive for the appearence of $S^{\text{op}}$ instead of $S$ is the way we compose functions. We have the following (which is easily verified):
If we regard the ring $S$ as a right module ($S_{S}$), then $\text{End}(S_{S}) \simeq S$ and if we regard $S$ as a left module ($_{S}S$), then $\text{End}(_{S}S) \simeq S^{\text{op}}$ (those are ring isomorphisms).
Now, for more details, let us examine the necessity of those propositions closely:
If $F : \text{Mod}_{S} \rightarrow \text{Mod}_{R}$ is an equivalence of categories, then putting $P_{R} = F(S_{S})$, we may show that $P_{R}$ is a progenerator, and to verify that $\text{End}(P_{R}) \simeq S$, we proceed like this: $$\text{End}(P_{R}) = \text{End}(F(S_{S})) \simeq \text{End}(S_{S}) \simeq S.$$
And for the left case we have:
If $F : \text{ }_{S}\text{Mod} \rightarrow \text{ }_{R}\text{Mod}$ is an equivalence of categories, then putting $_{R}P = F(_{S}S)$, we may show that $_{R}P$ is a progenerator, and we have: $$\text{End}(_{R}P) = \text{End}(F(_{S}S)) \simeq \text{End}(_{S}S) \simeq S^{\text{op}}.$$
To prove the sufficiency of the propositions, we proceed as follows:
If $P$ is a progenerator of $\text{Mod}_{R}$ such that $\text{End}(P) \simeq S$, then we may show that $$\text{Hom}_{R}(P,-) : \text{Mod}_{R} \rightarrow \text{Mod}_{S}$$ is an equivalence of categories, and for $M$ in $\text{Mod}_{R}$, we regard $\text{Hom}_{R}(P,M)$ as a right $\text{End}(P)$-module in the usual way, by composing functions, so that it becomes a right $S$-module.
For the left case we have:
If $P$ is a progenerator of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$, then we may show that $$\text{Hom}_{R}(P,-) : \text{ }_{R}\text{Mod} \rightarrow \text{ }_{S}\text{Mod}$$ is an equivalence of categories, and for $M$ in $_{R}\text{Mod}$, we regard $\text{Hom}_{R}(P,M)$ as a left $\text{End}(P)^{\text{op}}$-module in the usual way, by composing functions and making the necessary adjustments (putting the $^{\text{op}}$) so that everything works, and then it becomes a left $S$-module, since $\text{End}(P)^{\text{op}} \simeq (S^{\text{op}})^{\text{op}} = S$.