If $A, B$ are Morita equivalent, then $\underline{\text{Nat}}(I_A, I_A)\cong\underline{\text{Nat}}(I_B, I_B)$.

Your approach is correct, and in fact it's a special case of a more abstract category-theoretic observation. Suppose that $\mathcal{C}$ and $\mathcal{D}$ are equivalent categories, say $F \colon \mathcal{C} \to \mathcal{D}$ provides us with an equivalence with inverse $G$. Then I claim that I can find an isomorphism $F^* \colon \operatorname{Nat}(I_{\mathcal{D}},I_{\mathcal{D}}) \to \operatorname{Nat}(I_{\mathcal{C}},I_{\mathcal{C}})$ with inverse $G^*$. Here's what $F^*$ does. Suppose we start with a natural transformation $\beta \colon I_D \to I_D$. Then I define $F^* \beta$ to be the natural transformation $I_C \to I_C$ associating to an object $c$ the morphism $$c \to G(F(c)) \xrightarrow{G(\beta(F(c)))} G(F(c)) \to c$$ You need to check that this is well-defined. For instance, if you have a map $c \to c'$, then $$\require{AMScd} \begin{CD} c @>{}>> c'\\ @V{\beta(c)}VV @VV{\beta(c')}V\\ c @>>> c' \end{CD}$$ must commute. This may sound daunting, but when you write it out you'll find that it boils down to the naturality of the transformations $I \leftrightarrow G\circ F$ that commute from the fact that they're mutual inverses.

Through a symmetrical process one can find what $G^*$ should be.

Now we need to check that $F^* \circ G^* = I$ and $G^* \circ F^* = I$. This is going to be quite a trainwreck when writing it all out, but what you should always keep in mind is that, no matter how technical it seems, proofs in category theory always proceed by just 'following your nose'. Without further ado, let me try to unravel $G^* F^* \beta$ for a natural transformation $\beta$ and hope I don't make any mistakes. It is the natural map sending an object $d$ to the morphism $$d \to FG(d) \to FGFG(d) \xrightarrow{FG\beta FG(d)} FGFG(d) \to FG(d) \to d.$$ Proving that it's a natural transformation will ultimately rely simply on the naturality of the transformations $I \leftrightarrow FG$. Now to show that this natural transformation $G^* F^* \beta$ is just $\beta$ (up to isomorphism) note that the diagram $$\require{AMScd} \begin{CD} d @>{\beta(d)}>> d\\ @V{}VV @VV{}V\\ FG(d) @>>{\\beta (FG(d))}> FG(d)\\ @V{}VV @VV{}V\\ FGFG(d) @>>{FG\beta FG(d)}> FGFG(d) \end{CD}$$ commutes by all kinds of naturalities.

Note that the given data should already ensure such a canonical isomorphism $\varphi:N\to P\otimes_A M$, namely using the specific $M=Q\otimes_B N$ and the natural isomorphism $P\otimes_A(Q\otimes_B N)\cong N$. If you use that, everything should work as intended.

Nevertheless, as this ought to be the case for general equivalent categories as well, a more abstract approach could be more beneficial:

Suppose $F:\mathcal A\to\mathcal B$ and $G:\mathcal B\to\mathcal A$ are an equivalence pair of functors with adjoint natural isomorphisms $\eta:1_{\mathcal A}\to GF,\ \varepsilon:FG\to 1_{\mathcal B}$.
Then for a natural transformation $\lambda:1_{\mathcal A}\to 1_{\mathcal A}$ we can define $\overline\lambda$ by $$\overline\lambda_B:=\varepsilon_B\circ F(\lambda_{GB})\circ \varepsilon^{-1}_B\,,$$ and similarly you can define the other map ${\rm Nat}(1_{\mathcal B},1_{\mathcal B})\to{\rm Nat}(1_{\mathcal A},1_{\mathcal A})$ using $\eta$, and verify that they are inverses of each other.