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.