$\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\diag}{\operatorname{diag}}$ $\newcommand{\SL}{\operatorname{SL}}$ $\newcommand{\SO}{\operatorname{SO}_2}$ $\newcommand{\sig}{\sigma}$
This is a self-answered question. Alternative solutions are welcomed, of course.
Set $\Sigma=\begin{pmatrix} \sig & 0 \\\ 0 & \sig\end{pmatrix}$, $\sig >0$. Let $X \in \SL_2$ satisfy $\dist(\Sigma,\SL_2)=d(\Sigma,X).$ Lagrange's multipliers gives $$\Sigma= X-\lambda X^{-T}, \,\,\, \det X=1\tag{1}$$ for some real $\lambda$.
Claim: $(X,\lambda)$ form a solution to the system $(2)$ if and only if one of the following holds:
- $(X,\lambda)=(\id,1-\sig)$
- $(X,\lambda)=(-\id,1+\sig)$
- $\lambda=-1$, and $X= \begin{pmatrix} a & b \\ b & d \end{pmatrix}$ for some real $a,b,d$ satisfying $$\sigma = a+d \,\,\text{ and }\,\,ad-b^2=1.$$
We now prove that solutions of the last form exist if and only if $\sig \ge 2$, and that for $\sig=2$, it reduces to the other two previous cases $\pm \id$.
So, for $0<\sig \le 2$, there are exactly two (isolated) solutions $(X,\lambda)$, and for $\sig > 2$, every solution satsifies $\lambda=-1$, and the solutions form a disjoint union of two one-dimensional submanifolds, parametrised by $a$. (since $ b=\pm \sqrt{a(\sig-a)-1}$, $a+d=\sig$, $a$ determines $d$ and $b$ up to a sign).
Proof of $\sig \ge 2$:
Since $ad=b^2+1>0,\,\,\,$ $a,d$ have the same sign, and since $a+d=\sig>0$ this forces $a,d$ to be positive, so $$0<a,d<a+d=\sig. \tag{3}$$
By the AM-GM inequality, $$ \frac{\sig}{2}=\frac{a+d}{2} \ge \sqrt{ad} = \sqrt{b^2+1} \ge 1. $$
Now, $\sig=2$ holds if and only if $$ b=0, a=d, $$ so $X=a\id$, and $\det X=a^2=1$ reduces this to the other two cases $\pm \id$.
$\newcommand{\id}{\operatorname{Id}}$
Write $ X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $$ X-\lambda X^{-T} = \begin{pmatrix} a - md & b + mc \\ c + mb & d - ma \end{pmatrix} , \tag{1}$$ where $m=\lambda/\Delta$, and $\Delta: = ad-bc$.
Asking for the off-diagonal entries to be $0$ implies $b+mc=c+mb$, i.e. $$ (b-c)(1-m)=0, $$ so either $b=c$, or $m=1$.
We exclude the case $m=1$:
If $m=1$, then $$X-\lambda X^{-T}=\begin{pmatrix} a - d & b+c \\ b+c & d - a \end{pmatrix}, $$ which contradicts $X-\lambda X^{-T}=\Sigma$. (the diagonal elements of $X-\lambda X^{-T}$ are $a-d, -(a-d)$, which cannot both equal the positive $\sigma$.
We are left with the case where $b=c$. If $b=c=0$, then $X$ is diagonal, which we treat separately below.
If $b=c \neq 0$, then $0=b+mc=b(1+m)$, i.e. $m=-1$. Plugging this into equation $(1)$, we get $$ X-\lambda X^{-T}=(a+d)\id, $$ so $a+d=\sigma$.
We showed that $$ X = \begin{pmatrix} a & b \\ b & d \end{pmatrix}, $$ where $a+d=\sigma$. Furthermore, we have $$ \Delta=\det X=ad-b^2=1, $$ so $-1=m=\lambda$, $b \neq 0$.
We turn to treat the diagonal case, where $X=\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$.
Equation $(1)$ then becomes $$ X-\lambda X^{-T} = \begin{pmatrix} a - md & 0 \\ 0 & d - ma \end{pmatrix}, $$ so $X-\lambda X^{-T}=\Sigma$ implies $$ a-md=d-ma=\sigma \tag{2}, $$ i.e. $ (a-d)(1+m)=0. \tag{3} $ If $a=d$, then $\det X=a^2=1$, so $a=\pm \id$, $X=\pm \id$. Now, $m=\lambda$, we get $a(1-\lambda)=\sigma$. Thus:
If $a=1$: $X=\id,1-\lambda=\sigma$.
If $a=-1$: $X=-\id,\lambda-1=\sigma$.
We return to equation $(3)$. We anlaysed the case $a=d$. If $a \neq d$, then $m=\lambda=-1$, so equation $(2)$ implies $a+d=\sigma$. Thus $X=\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$, where $a+d=\sigma$. This is a special case of form $3$ in the question where $b=0$.