I am back with a good new: you have nothing to do because there are (in general) no critical points! It is the case if, in particular, $S$ is invertible. Recall that $\mathcal{L}(A)=log|I+AS|-q^T(I+AS)^{-1}Aq$ and (according to my previous post) $D\mathcal{L}_A=\mathcal{L}'(A)=H\rightarrow tr((I+AS)^{-1}HS)-q^T(-(I+AS)^{-1}HS(I+AS)^{-1}A+(I+AS)^{-1}H)q$.

Thus $D\mathcal{L}_A=0$ is equivalent to: for any matrix $H$, $tr(S(I+AS)^{-1}H)=q^T(-(I+AS)^{-1}H(I-S(I+AS)^{-1}A)q$, that is in the form

(*) $tr(UH)=r^THs$ where $r,s$ are vectors and with $U=S(I+AS)^{-1}$.

Lemma: If (*) is true for any $H$, then necessarily $rank(U)\le 1$.

Proof: for any $h_{i,j}$, $\sum_{i,j}u_{j,i}h_{i,j}=\sum_{i,j}r_is_jh_{i,j}$. Thus $u_{j,i}=s_jr_i$, that is $U=sr^T$. Therefore $rank(U)\leq 1$.

Conclusion: Here $A,S\geq 0$; thus $I+AS$ is invertible and consequently, $\mathcal{L}$ admit some critical points only if $rank(S)\leq 1$, that is $S=ww^T$ where $w$ is a vector.

I am back with good news: you have nothing to do because there are (in general) no critical points! It is the case if, in particular, $S$ is invertible. Recall that $\mathcal{L}(A)=log|I+AS|-q^T(I+AS)^{-1}Aq$ and (according to my previous post) $D\mathcal{L}_A=\mathcal{L}'(A)=H\rightarrow tr((I+AS)^{-1}HS)-q^T(-(I+AS)^{-1}HS(I+AS)^{-1}A+(I+AS)^{-1}H)q$.

Thus $D\mathcal{L}_A=0$ is equivalent to: for any matrix $H$, $tr(S(I+AS)^{-1}H)=q^T(-(I+AS)^{-1}H(I-S(I+AS)^{-1}A)q$, that is in the form

(*) $tr(UH)=r^THs$ where $r,s$ are vectors and with $U=S(I+AS)^{-1}$.

Lemma: If (*) is true for any $H$, then necessarily $rank(U)\le 1$.

Proof: for any $h_{i,j}$, $\sum_{i,j}u_{j,i}h_{i,j}=\sum_{i,j}r_is_jh_{i,j}$. Thus $u_{j,i}=s_jr_i$, that is $U=sr^T$. Therefore $rank(U)\leq 1$.

Conclusion: Here $A,S\geq 0$; thus $I+AS$ is invertible and consequently, $\mathcal{L}$ admits some critical points only if $rank(S)\leq 1$, that is $S=ww^T$ where $w$ is a vector.