(Basically) Full answer
- For $s \geq 1$ we always take the matrix with diagonals $s^{1/n}$.
- For $s < 1$ and for $n$ large enough enough we have two possibilities. For $s$ large enough we take (still) the matrix with diagonals $s^{1/n}$. For $s$ small enough we instead solve $$a(1-a)^{n-1} = s$$ and take the smaller of the (at most two) solutions. The optimizer will be the matrix with one diagonal $a$ and the rest $1-a$. (Note for $s$ small enough we will have $a \approx s$ and $dist \approx 1$.
Intuitive optimizers
First, to get some intuition for the problem realize that the behavior of the solution is going to depend on $s$. If $s$ is very large, then it is best to take all the eigenvalues equal. This is easy to visualize: the closest point on the graph of $xyz=1000$ to $(1,1,1)$ is $(10,10,10)$.
On the other hand, if $s$ is very small then taking all the values the same gives a distance of about $n$. By taking one eigenvalue to be $s$ and the rest to be $1$ we can have a distance of $1$ from the identity.
Dimensionless problem
We pick up from your optimization problem (1). It is convenient to write $\alpha = k/n$ and $L = \log(s)/n$ so that your optimization problem is equivalent to minimizing $$G(a,b,\alpha) = \alpha (a-1)^2 + (1-\alpha) (b-1)^2$$ over the set $$H(a,b,\alpha) = L, \quad a>0,\quad b>0,\quad \alpha \in [0,1], \quad n\alpha \in \mathbb{N}$$ where $H(a,b,\alpha)$ is $$H(a,b,\alpha) = \alpha \log (a) + (1-\alpha) \log (b).$$
We can perform the method of Lagrange multipliers in the $a$ and $b$ coordinates. This will give us that either $b=a$ or $b = (1-a)$. A more direct path to this result is to observe in your equation $$\sigma^2 - \sigma - \lambda = 0 \text{ for some } \lambda \in \mathbb{R}$$ we have that if $\sigma$ is a solution, so is $(1-\sigma)$.
The easy competitor
Now if $a = b$ then $\alpha$ is irrelevant, and we have the first competitor to the minimization $$a = b = e^L, \quad \alpha = \text{anything}, \quad dist = (e^L-1)^2$$
The harder competitor The case $b = 1-a$ is harder to analyze. First note that for this solution we must restrict to $a < 1$. Given that $b = 1-a$ we can rewrite our optimization as minimizing $$\alpha (a-1)^2 + (1-\alpha)a^2 $$ over the set \begin{equation}\tag{1} \alpha \log(a) + (1 - \alpha) \log(1-a) = L \end{equation} \begin{equation} \tag{2} a \in (0,1/2), \quad a \leq \min(e^L, 1-e^L), \quad n\alpha \in \mathbb{N} \end{equation} The first constraint above comes from assuming (WLOG) $a$ is smaller than $b$. The second is from observing that $L$ is a convex combination of $\log a$ and $\log (1-a)$ so $\log (a) \leq L \leq \log(1-a)$.
To simplify, we can solve the constraint $\alpha \log a + (1-\alpha)\log (1-a)$ for $\alpha$: $$\alpha = \frac{L - \log(1-a)}{\log(a) - \log(1-a)}$$ and now rewrite our minimization as: minimize $$f(a) = \frac{(L - \log(1-a))(1-a)^2 + (\log(a)-L)a^2}{\log(a)-\log(1-a)}$$ over $$a \in (0, \min(e^L, 1-e^L)), \quad n\alpha \in \mathbb{N}$$
Now we claim the following three facts:
- $f(a)$ as zero or one critical points
- $\lim_{a \to 0}f(a) = 0$
- $f(e^L) = f(1-e^L) = (e^{L}-1)^2$
Given these three facts we see that if we forget the condition $n \alpha \in \mathbb{N}$ from our constraints we see that the infimum of $f$ is $0$ which is not realized for $a > 0$. The function either looks like
or
(These are L = -1/5 and -3.) The red line in these pictures is the value from taking the matrix of diagonals, (e^L - 1)^2.
If we now enforce the discrete condition $n \alpha \in \mathbb{N}$ we see that it suffices to check the smallest $a$ possible against $(e^L -1 )^2$.
Checking these facts I checked items 2 and 3 with a CAS. For item 1 I did the following. First implicitly differentiate the constraint (1) with respect to $a$ to find $$ \frac{d\alpha}{da} \left(\log a - \log(1-a) \right) = - \left(\frac{\alpha}{a} - \frac{1-\alpha}{1-a} \right) $$ Then differentiate $f$ and set it to zero to find $$ 2 (a - \alpha) \left(\log a - \log(1-a) \right) = (-2a + 1) \left( \frac{\alpha}{a} - \frac{1-\alpha}{1-a} \right) $$ Multiply by $a(1-a)$ to find $$ 2a(1-a)(a-\alpha) \left( \log a - \log(1-a) \right) = (-2a+1) \left(\alpha - a \right) $$ If $\alpha \neq a$ we can divide by $\alpha-a$ and find (checking that there's only one solution) $a = 1/2$. This is disallowed by our constraints (2). On the other hand, we have a solution if $\alpha = a$. Then, check that there is at most one solution to (1) with $\alpha = a$.