**(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 $(10,10,10)$.
[![enter image description here][1]][1]

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:

 1. $f(a)$ as zero or one critical points
 2. $\lim_{a \to 0}f(a) = 0$
 3. $f(e^L) = f(1-e^L) = (e^{L}-1)^2$

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
[![enter image description here][2]][2]
or
[![enter image description here][3]][3]
(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 $\alpha = a$.


  [1]: https://i.sstatic.net/3aLlh.png
  [2]: https://i.sstatic.net/cp8fk.png
  [3]: https://i.sstatic.net/8BTIT.png