we will view $S_{\tau}$ as an operator acting on $L^2[s,\infty)$ in what follows.\
So the problem becomes to show that, as $n,N {\rightarrow}\infty$ $$\label{whattoshow1}
\phi_{\tau}(s),\psi_{\tau}(s){\rightarrow}\frac{1}{\sqrt{2}}\mathrm{Ai}(s) \; ,$$ and that $\forall s_0 \in \mathbb{R}$, there exists $N_0(s_0)$ such that if $N>N_0$, we have on $[s_0,\infty)$, $$\label{whattoshow2}
\phi_{\tau}(s),\psi_{\tau}(s)=O({\mathrm{e}}^{-s/2}) \; .$$ Once this is shown (we give more details on this later), we can show that $S_{\tau} {\rightarrow}\bar{S}$ in the trace class norm of operators on $L^2[s,\infty)$. A classical way to do it is described in the remark at the end of section 3 of @imj, which bounds the trace class norm of the difference of $S_{\tau}-\bar{S}$ in terms of the Hilbert-Schmidt norm of operators whose kernels are related to $\phi_{\tau},
\psi_{\tau}$ and ${\mathrm{Ai}}$. This leads to the conclusion that $$\det({\mathrm{Id}}-S_{\tau}) {\rightarrow}\det({\mathrm{Id}}-\bar{S}) \;,$$ since $\det$ is continuous with respect to trace class norm. Therefore, the largest eigenvalue of $X^*X$ has the behavior it was claimed it has.

Real Case
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In the real case, using arguments from @tw96 and @widom99, @imj gets a representation similar to (\[freddet\]), this time involving an operator with kernel a $2\times 2$ matrix (instead of scalar in the complex case). He is then able to relate it to 