Entropy change when indistinguishable particles suddenly become distinguishable
Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.

Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation

$$\begin{align}
Q_\mathrm{tr,indis} &amp;= \frac{q_\mathrm{tr}^N}{N!} \\
\ln Q_\mathrm{tr,indis} &amp;= N \ln q_\mathrm{tr} - \ln(N!) \\
&amp;\approx N \ln q_\mathrm{tr} - N \ln N + N \\
&amp;= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right)
\end{align}$$

Plugging $N = 1$ into Stirling's approximation however tells you

$$\ln (1!) \approx 1\ln 1 - 1$$

or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!

Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result

$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$

which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.