Use the standard relationship between acceleration in the two frames of reference.
i.e. the proper acceleration $a$ is given by
$$ a = \gamma^3 \frac{dv}{dt}, \ \ \ \ \ \ \ \ \gamma = \frac{dt}{d\tau} $$
$$ \frac{dv}{d\tau} = \frac{dv}{dt} \frac{dt}{d\tau} = \gamma^{-2} a = (1-v^2)a$$
This can be integrated to give $v$ and hence $\gamma$ as a function of $\tau$.
$$ \int \frac{dv}{1-v^2} = \int a\ d\tau$$
Let $v = \tanh(x)$ and use the identity $1 - \tanh^2(x) = 1/\cosh^2(x)$
$$\frac{dv}{dx} = \frac{\cosh^2(x) - \sinh^2(x)}{\cosh^2(x)} = \frac{1}{\cosh^2(x)}$$
and so the integral becomes
$$ \int dx = \int a\ d\tau$$
$$ \tanh^{-1} (v) = a\tau + A,$$
where $A$ is a constant determined by the initial velocity.
Let $v=v_0$ when $\tau=0, $hence:
$$ v = \tanh[a\tau + \tanh^{-1}(v_0)]$$
The doppler shift can be written as:
$$ \omega = \omega_0 (1-v)\gamma$$
NB: This expression comes from here, with the source at rest, but is I think only strictly valid when the velocity of the observer does not change significantly between wavefronts. For optical light, this requires that (expressing $a$ in SI units for a moment) $a \ll 10^{24}$ ms$^{-2}$ - which is probably ok for a spaceship!
$$ \omega = \omega_0\left[1 - \tanh[a\tau + \tanh^{-1}(v_0)]\right]\left[1 - \tanh^2[a\tau + \tanh^{-1}(v_0)]\right]^{-1/2}$$
$$ \omega = \omega_0 \left[1 - \tanh[a\tau + \tanh^{-1}(v_0)]\right]\cosh[a\tau + \tanh^{-1}(v_0)]$$.
This is the general expression. For the specific case addressed by the OP, we have $v_0=0$. In this case:
$$\omega = \omega_0[1 - \tanh(a\tau)]\cosh(a\tau)$$
$$\omega = \omega_0\left[\frac{\cosh(a\tau) - \sinh(a\tau)}{\cosh(a\tau)}\right] \cosh(a\tau)$$
Expressing the hyperbolic functions in terms of exponentials:
$$\omega = \frac{\omega_0}{2}[\exp(a\tau) + \exp(-a\tau) - \exp(a\tau) + \exp(-a\tau)] = \omega_0 \exp(-a\tau)$$
as required.
A similar treatment is provided by Cochran 1989 (section II), leading to the same result.
A more useful result is obtained by noting that a coordinate transform of the form
$$ \tau^{\prime} = \tau + \frac{\tanh^{-1}(v_0)}{a}$$
can make life more easy for general cases, since this also leads to the result
$$ \omega = \omega_0 \exp(-a\tau^{\prime})$$
This does make life easier - for instance we can show that we recover the standard doppler shift when $a=0$, since $a\tau^{\prime} = \tanh^{-1}(v_0)$ and so
$$\omega = \omega_0 \exp[-\tanh^{-1}(v_0)] = \omega_0 \exp\left[-\frac{1}{2}\ln \left(\frac{1+v_0}{1-v_0}\right)\right]$$
$$ \omega = \omega_0\left( \frac{1+v_0}{1-v_0}\right)^{-1/2} = \omega_0(1-v_0)[(1+v_0)(1-v_0)]^{-1/2} = \omega_0(1-v_0)\gamma\ .$$
