As you have already suspected: you cannot simply say $\tau(s) = s$.
I think you might be missing some theoretical background of how the geodesic equations are often derived.
I am going to use a bit of matrix notations, to skip some of the indexing. So $$x = \begin{bmatrix} x^i\end{bmatrix} = \begin{bmatrix} x^0\\x^1\\x^2\\x^3\end{bmatrix} \, \text{ and } \, g(x) = \Big[g_{ij}(x)\Big]_{i,j = 0}^{3} \, \text{ is the 4 by 4 metric tensor describing gravity} $$
The description of the motion of a point-mass particle in General Relativity starts with the relativistic Lagrangian $$L = -m\,\sqrt{- \, \dot{x}^T\,g(x)\, \dot{x}}$$ where $\dot{x} = \cfrac{dx}{d\lambda}$. With its help, one can define the action
$$S[\gamma] = -m\, \int_{\lambda_1}^{\lambda_2} \, \sqrt{- \, \dot{x}^T(\lambda)\,g\big(x(\lambda)\big)\, \dot{x}(\lambda)}\,d\lambda $$
where $$\gamma = \{x(\lambda) \, : \, \lambda \in [\lambda_1, \lambda_2]\}$$ can be any smooth time-like curve in 4D space-time, parametrized by some arbitrary parameter $\lambda$ and connecting two fixed space-time points, i.e. $x(\lambda_1) = x_1$ and $x(\lambda_2) = x_2$ are fixed with $\lambda_1$ and $\lambda_2$ also fixed (this is a very important conceptual requirement that is used in the derivation of the Euler-Lagrange equations!). Fundamentally, in the philosophy of General Relativity, this parameter $\lambda$ is not of any importance to the theory. Only the geometric shape of the curve $\gamma$ in 4D space-time matters and not the specific parametrization $\lambda$. After all, this curve $\gamma = \{x(\lambda)\}$ is supposed to be a space-time time-like smooth curve for which the action functional $S[\gamma]$ attains a critical value (i.e. this is the definition of a time-like geodesic). This is a geometric property independent of any parametrization $\lambda$, so we really care about the geodesic $\gamma$ as a geometric curve and not as a parametrized curve, so we are free to change the parametrization as we see fit in order to simplify our calculations.
The motion of a mass-particle in General Relativity is a time-like geodesic. As per the definition of a geodesic, we need to look for the critical (non-parametrized!!!) curves that connect two fixed space-time points $x_1$ and $x_2$ and that satisfy the critical value condition
$$\delta S[\gamma] = 0$$
In coordinates $[x^i]$ and with respect to a generic parametrization $\lambda$, the equation $\delta S[\gamma] = 0$ is equivalent to the Euler-Lagrange differential equations
$$\frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{2\, \sqrt{-\, \dot{x}^T\,g(x)\,\dot{x}\,}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ where
$$ \dot{x}^T\, \frac{\partial g}{\partial x}(x)\, \dot{x}\, = \,
\begin{bmatrix}
\frac{\partial g_{ij}}{\partial x^0}(x)\,\dot{x}^i\,\dot{x}^j \\
\frac{\partial g_{ij}}{\partial x^1}(x)\,\dot{x}^i\,\dot{x}^j \\
\frac{\partial g_{ij}}{\partial x^2}(x)\,\dot{x}^i\,\dot{x}^j \\
\frac{\partial g_{ij}}{\partial x^3}(x)\,\dot{x}^i\,\dot{x}^j
\end{bmatrix}$$ for short. Take a solution (time-like) $\gamma = \{ x(\lambda)\, : \, \lambda \}$ of the Euler-Lagrange equations above. As I have already emphasized, the parametrization of $\gamma$ with respect to $\lambda$ is not important for us. Therefore, I can define the function
$$s = s(\lambda) = \int_{\lambda_0}^{\lambda}\, \sqrt{-\, \dot{x}(\zeta)^T \, g\big(\, x(\zeta)\,\big)\, \dot{x}(\zeta)\,}\, d\zeta$$
with derivative
$$\frac{ds}{d\lambda} = \sqrt{-\, \dot{x}(\lambda)^T \, g\big(\, x(\lambda)\,\big)\, \dot{x}(\lambda)\,} \, > \,0$$ Thus the function $s = s(\lambda)$ is strictly increasing and therefore invertible, i.e. there is $\lambda = \lambda(s)$. Consequently, we can re-parametrize our solution curve $\gamma$ as $$\gamma = \{ \, x(s) \, : \, s \, \} \, \text{ where } \, x(s)= x\big(\lambda(s)\big)$$ Observe that
$$\gamma = \{\,x(s)\, : \, s \,\} = \{\, x(\lambda)\, : \, \lambda \, \}$$ in other words, this is the same curve in space time, but parametrized in two different ways. Denote $x' = \frac{dx}{ds}$. Furthermore,
$$x' = \frac{dx}{ds} =\frac{d\lambda}{ds} \frac{dx}{d\lambda} = \left( \frac{ds}{d\lambda}\right)^{-1} \frac{dx}{d\lambda} = \frac{1}{\sqrt{- \, \dot{x}^T \, g(x) \, \dot{x}}\,}\, \frac{dx}{d\lambda}$$
and in particular $$\frac{d}{ds} = \frac{1}{\sqrt{- \, \dot{x}^T \, g(x) \, \dot{x}}\,}\, \frac{d}{d\lambda} $$ Recall that the curve $\gamma$ is a critical curve for the action $S[\gamma]$, i.e. $\delta S[\gamma] = 0$. When $\gamma$ is parametrized with respect to $\lambda$, it's coordinate parametrization $\gamma = \{\, x(\lambda) \, : \, \lambda\}$ solves the Euler-Lagrange equations
$$\frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{2\, \sqrt{-\, \dot{x}^T\,g(x)\,\dot{x}\,}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ whose both sides I can multiply by $\frac{1}{\sqrt{-\, \dot{x}^T\, g(x) \, \dot{x}}\,}$ and obtain the equivalent equations
$$\frac{1}{\sqrt{-\, \dot{x}^T\, g(x) \, \dot{x}}\,} \, \frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{-\, 2\, \dot{x}^T\,g(x)\,\dot{x}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ It is easy to check that with the new parametrization $\gamma = \{\, x(s) \, : \, s\, \}$ $$\sqrt{-\, \frac{dx}{ds}^T \, g(x) \, \frac{dx}{ds}} =\sqrt{-\, (x')^T \, g(x) \, x'} = 1$$ Consequently, after the reparametrization $\lambda = \lambda(s)$ the Euler-Lagrange equations turn into the equivalent simplified equations
$$\frac{d}{ds}\left(\, m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\,\left( \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right) $$ which is solved by $\gamma = \{\, x(s)\, : \, \tau\,\}$.
In other words we have proven that any solution $\gamma$ to the original Euler-Lagrange equations, after the appropriate reparametrization, solves the simplified Euler-Lagrange equations. This means that a curve $\gamma$ is a critical curve of the action $S[\gamma]$, i.e. $\delta S[\gamma] = 0$ if and only if it solves the simplified Euler-Lagrange differential equations
$$\frac{d}{ds}\left(\,m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\, \left(\, \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right) $$ where the resulting parametrized solution $\gamma = \{\, x(s)\, : \, s\,\}$ is paremtrized with respect to proper time, i.e. $\sqrt{ - \, x'(s)^T\, g\big(x(s)\, x'(s)\big)} = 1$ for any $s$.
It is very important to note that the simplified Lagrange differential equations
$$\frac{d}{ds}\left(\,m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\, \left(\, \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right)$$
happen to match the Euler-Lagrange equations coming from the Lagrangian
$$L = \frac{m}{2}\,\frac{dx}{ds}^T g(x) \frac{dx}{ds}$$ but this latter Lagrangian is conceptually not well defined in the framework of General relativity, as it is not independent on reparametrization. So, it is probably safer to avoid it in the framework of General Relativity, as it can lead to wrong results. This Lagrangian is not a problem in the theory of classical Lagrangian mechanics,
but this is because there time is an absolute variable, separate from space and there is no need for treating time and space on equal footing.
Coming back to the problem at hand, there are various ways one can calculate the geodesics of 1,1 de Sitter space. One way is to think of the 1,1 de Sitter space as a one-sheeted hyperboloid embedded in 2, 1 Minkowski space, invariant under the action of the Lorentz group, and by using a symmetry argument, to show that the geodesics are the curves obtained by the intersection of 2D planes through the origin with the de Sitter hyperboloid. Another approach could be to observe that the Lagrangian $$L = -m\sqrt{\left(\frac{d\tau}{d\lambda}\right)^2 - \,\cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2}$$ does not depend explicitly on the variable $\phi$, which means the Lagrangian is invariant under the action of the one-parametr group of symmetries $\phi \mapsto \phi + \sigma$ so one can apply Noether's theorem and obtain
$$\frac{\partial }{\partial \dot{\phi}} \sqrt{\left(\frac{d\tau}{d\lambda}\right)^2 - \,\cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2 } \, = \, c_0 \,(= \text{const})$$
then perform the differentiation in the latter identity and after that express $\phi = \phi(\tau)$ as a function of $\tau$. However, I think all of these methods are not first principle methods, so I would just use the direct definitions of geodesics.
Let's see how the theoretical procedure outlined above applies to your example. In your case, we can assume that the time-like geodesics $\big(\tau = \tau(s), \, \phi = \phi(s)\big)$ are parametrized by the proper time $s$ and thus, satisfy the simplified Euler-Lagrange equations:
\begin{align}
&\frac{d}{ds} \Big( -\, \frac{d\tau}{ds}\Big) \, = \, \frac{1}{2} \frac{\partial}{\partial \tau} \left( - \, \Big(\, \frac{d\tau}{ds}\,\Big)^2 \, + \, \cosh^2(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\right)\\
&\frac{d}{ds} \Big( \,\cosh^2(\tau)\, \frac{d\phi}{ds}\Big) \, = \, \frac{1}{2} \frac{\partial}{\partial \phi} \left( - \, \Big(\, \frac{d\tau}{ds}\,\Big)^2 \, + \, \cosh^2(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\right)
\end{align}
After performing most of the differentiations, we obtain the system
\begin{align}
& -\, \frac{d^2\tau}{ds^2} \, = \, \cosh(\tau)\sinh(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\\
&\frac{d}{ds} \Big( \,\cosh^2(\tau)\, \frac{d\phi}{ds}\Big) \, = \, 0
\end{align}
The second equation can be immediately integrated once with respect to $s$, yielding the system
\begin{align}
& -\, \frac{d^2\tau}{ds^2} \, = \, \cosh(\tau)\sinh(\tau)\, \Big(\, \frac{d\phi}{ds}\,\Big)^2\\
&\cosh^2(\tau)\, \frac{d\phi}{ds} \, = \, c_0
\end{align}
where $c_0$ is a constant. Thus
\begin{align}
& \frac{d^2\tau}{ds^2} \, = \, -\, \cosh(\tau)\sinh(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align}
One can plug the second equation in the first
\begin{align}
& \frac{d^2\tau}{ds^2} \, = \, -\, \cosh(\tau)\sinh(\tau)\Big(\, \frac{c_0}{\cosh^2(\tau)}\,\Big)^2 \\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align}
and simplify
\begin{align}
& \frac{d^2\tau}{ds^2} \, = \, -\, c_0^2\,\frac{\sinh(\tau)}{\cosh^3(\tau)}\\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align}
The first equation decouples from the second because it is an equation only for the variable $\tau$. If you multiply both sides of the first equation by $\frac{d\tau}{ds}$ then you can integrate it once and obtain
\begin{align}
& \frac{1}{2}\left(\frac{d\tau}{ds} \right)^2 \, = \, \frac{c_1}{2} \, -\, c_0^2\, \int \frac{\sinh(\tau)}{\cosh^3(\tau)} d\tau\\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align} After integrating the right-hand side, the equations become
\begin{align}
& \frac{1}{2}\left(\frac{d\tau}{ds} \right)^2 \, = \, \frac{c_1}{2} \, + \, \frac{c_0^2}{2\,\cosh^2(\tau)}\\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align} divide by two and take a square root
\begin{align}
&\frac{d\tau}{ds} \, = \, \pm \sqrt{c_1 \, + \, \frac{c_0^2}{\cosh^2(\tau)} \, }\\
&\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)}
\end{align}
From here, if you decide to express the proper time $s = s(\tau)$ as a function of $\tau$, then the variable $\phi = \phi(s) = \phi\big(s(\tau)\big)$ becomes also a function of $\tau$ and by the chain rule $$\frac{d\phi}{d\tau} = \frac{d\phi}{ds} \, \frac{ds}{d\tau} = \frac{\,\,\left( \frac{d\phi}{ds}\right)\,\,}{\left(\frac{d\tau}{ds}\right)} = \, \pm\,\frac{\,\,\left( \frac{c_0}{\cosh^2(\tau)}\right)\,\,}{\left(\sqrt{c_1 \, + \, \frac{c_0^2}{\cosh^2(\tau)} \, }\right)}$$ so we end up with a differential equation that is readily integrable
$$\frac{d\phi}{d\tau} = \,\frac{\left(\frac{c_0}{\sqrt{c_1}\,\cosh^2(\tau)}\right)}{\sqrt{1 \, + \, \frac{c_0^2}{c_1\, \cosh^2(\tau)} \, }}$$ where I have taken only the plus version of the equation. Consequently,
$$\phi = \phi_0 \, + \, \int \, \frac{\left(\frac{c_0}{\sqrt{c_1}\,\cosh^2(\tau)}\right)}{\sqrt{1 \, + \, \frac{c_0^2}{c_1\, \cosh^2(\tau)} \, }}\, d\tau$$
to solve the integral, you can for example set $a_0 = \frac{c_0}{\sqrt{c_1}}$ and perform the substitution
$$z = a_0\,\tanh(\tau) \,\, , \,\,\,\,\, dz = \frac{a_0}{\cosh^2(\tau)} \, d\tau$$
$$\frac{a_0^2}{\cosh^2(\tau)} = a_0^2\,\frac{\cosh^2(\tau) - \sinh^2(\tau)}{\cosh^2(\tau)} = a_0^2 - a_0^2\,\tanh^2(\tau) = a_0^2 - z^2$$ so the integral becomes
$$\phi = \phi_0 \, + \, \int \, \frac{1}{\sqrt{(1 \, + \, {a_0^2}) \,-\, z^2}}\, dz = \arcsin\left(\frac{z}{\sqrt{1+a_0^2}}\right)$$ Thus, finally we find the formula
$$\phi = \phi \pm \arcsin\big(\,k_0\tanh(\tau)\,\big)$$
where $k_0 = \frac{a_0}{\sqrt{1 + a_0^2}}$ is a constant.
