Here are some thoughts of my own which derive from Will Sawin's answer (which solved the problem in the case $(A_1,\ldots,A_k) \in SL_2(\mathbb{R})^k$). Let us suppose that $(A_1,\ldots,A_k) \in SL_2^\pm(\mathbb{R})$ is a critical point of $\tau$ such that $\tau(A_1,\ldots,A_k)=0$. I claim that for every $r \in \{1,\ldots,k\}$, the equation $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left((A_{i_n}\cdots A_{i_{\ell+1}})X(A_{i_n}\cdots A_{i_{\ell+1}})^{-1}+(A_{i_\ell}\cdots A_{i_{1}})^{-1}X(A_{i_\ell}\cdots A_{i_{1}}))\right)=0$$ must be satisfied for all traceless matrices $2 \times 2$ real matrices $X$. Furthermore this is a necessary and sufficient condition for a zero $(A_1,\ldots,A_k)$ of $\tau$ to be a critical point of $\tau$.

To see this let us fix $r$ and $X$ and define $\sigma:=\det A_{i_n}\cdots A_{i_1} \in \{\pm1\}$. We suppose that $(A_1,\ldots,A_k)$ is both a zero and a critical point of $\tau$. If $(B_1,\ldots,B_k)$ is in the same connected component of $SL_2^\pm(\mathbb{R})^k$ as $(A_1,\ldots,A_k)$ then by the Cayley-Hamilton theorem $$(B_{i_n}\cdots B_{i_1})^2 - (\tau(B_1,\ldots,B_k))B_{i_n}\cdots B_{i_1} +\sigma I = 0.$$ Let us differentiate this equation along the one-parameter family $t \mapsto (A_1,\ldots,e^{tX}A_r,\ldots,A_k)$ at the value $t=0$. Since by hypothesis $\tau$ and its derivative are zero at $(A_1,\ldots,A_k)$, the only remaining terms are those arising from the first term, namely $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)A_{i_n}\cdots A_{i_1}$$ and $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} A_{i_n}\cdots A_{i_1}\left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)$$ and the total of these two sums must equal zero. Again by the Cayley-Hamilton theorem we have $$(A_{i_n}\cdots A_{i_1})^2=-\sigma I$$ so in particular $$A_{i_n}\cdots A_{i_1}A_{i_n}\cdots A_{i_{\ell+1}}=-\sigma (A_{i_\ell}\cdots A_{i_1})^{-1}$$ and $$A_{i_\ell}\cdots A_{i_1}A_{i_n}\cdots A_{i_1}=-\sigma (A_{i_n}\cdots A_{i_{\ell+1}})^{-1}.$$ Substituting these identities into the results obtained thus far we obtain the claimed identity. Note that all the steps of this argument are reversible.

Here are some thoughts of my own which derive from Will Sawin's answer (which solved the problem in the case $(A_1,\ldots,A_k) \in SL_2(\mathbb{R})^k$). Let us suppose that $(A_1,\ldots,A_k) \in SL_2^\pm(\mathbb{R})$ is a critical point of $\tau$ such that $\tau(A_1,\ldots,A_k)=0$. I claim that for every $r \in \{1,\ldots,k\}$, the equation $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left((A_{i_n}\cdots A_{i_{\ell+1}})X(A_{i_n}\cdots A_{i_{\ell+1}})^{-1}+(A_{i_\ell}\cdots A_{i_{1}})^{-1}X(A_{i_\ell}\cdots A_{i_{1}}))\right)=0$$ must be satisfied for all traceless matrices $2 \times 2$ real matrices $X$. Furthermore this is a necessary and sufficient condition for a zero $(A_1,\ldots,A_k)$ of $\tau$ to be a critical point of $\tau$.

To see this let us fix $r$ and $X$ and define $\sigma:=\det A_{i_n}\cdots A_{i_1} \in \{\pm1\}$. We suppose that $(A_1,\ldots,A_k)$ is both a zero and a critical point of $\tau$. If $(B_1,\ldots,B_k)$ is in the same connected component of $SL_2^\pm(\mathbb{R})^k$ as $(A_1,\ldots,A_k)$ then by the Cayley-Hamilton theorem $$(B_{i_n}\cdots B_{i_1})^2 - (\tau(B_1,\ldots,B_k))B_{i_n}\cdots B_{i_1} +\sigma I = 0.$$ Let us differentiate this equation along the one-parameter family $t \mapsto (A_1,\ldots,e^{tX}A_r,\ldots,A_k)$ at the value $t=0$. Since by hypothesis $\tau$ and its derivative are zero at $(A_1,\ldots,A_k)$, the only nonzero terms in the derivative of the Cayley-Hamilton identity are those arising from the first term, namely $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)A_{i_n}\cdots A_{i_1}$$ and $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} A_{i_n}\cdots A_{i_1}\left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)$$ and the total of these two sums must equal zero. Again by the Cayley-Hamilton theorem we have $$(A_{i_n}\cdots A_{i_1})^2=-\sigma I$$ so in particular $$A_{i_n}\cdots A_{i_1}A_{i_n}\cdots A_{i_{\ell+1}}=-\sigma (A_{i_\ell}\cdots A_{i_1})^{-1}$$ and $$A_{i_\ell}\cdots A_{i_1}A_{i_n}\cdots A_{i_1}=-\sigma (A_{i_n}\cdots A_{i_{\ell+1}})^{-1}.$$ Substituting these identities into the results obtained thus far we obtain the claimed identity. Note that all the steps of this argument are reversible.