An operator $\exp \left(\tau \frac{\partial}{\partial t}\right)$ when acting on a function $f(x,t)$ can be show to satisfy the following equation:
$$ \exp \left(\tau \frac{\partial}{\partial t}\right) f(x,t) = f(x,t+\tau)$$
This can be done by expanding the exponential operator as a Taylor series $(1 + \tau \frac{\partial}{\partial t} + \frac{\tau^2}{2!}\frac{\partial^2}{\partial t^2} + ...)$ and applying it on the function $f$ to give $\left(f(x,t) + \tau\frac{\partial}{\partial t} f(x,t)+ \frac{\tau^2}{2!}\frac{\partial^2}{\partial t^2} f(x,t) + ... \right)$. This is simply the Taylor expansion of $f(x,t+\tau)$.
If however, the function $f$ is a product of two function $g=g(x,0)$ and $h=h(x,0)$, would the following relation hold?
$$ \exp \left(\tau \frac{\partial}{\partial t}\right) \left[g(x,t) h(x,t)\right] = g(x,t+\tau) h(x,t+\tau)$$
How would I go about proving or disproving this?
As has been discussed in the comments, a function is still a function when you write it as a product. If you've proved a result for all functions, then it holds for all functions, including functions that you happen to write as a product. Note that all functions can be written as a product: $f(x)=g(x)h(x)$ with $g(x)\equiv1$ and $h(x)=f(x)$.
There are two errors in your presentation and derivation of the exponential operator result. First, $f(x,0)$ doesn't depend on $t$, so $\exp\left(t\frac{\partial}{\partial t}\right)$ annihilates it. Second, $t$ and $\frac{\partial}{\partial t}$ don't commute, so you can't reorder them like you did.
The correct relationship is
$$ \exp\left(\tau\frac{\partial}{\partial t}\right)\,f(x,t)=f(x,t+\tau)\;. $$
Edit:
Taking the long way home:
\begin{eqnarray*} \exp \left( \tau \frac{\partial}{\partial t} \right) (gh) &=&\sum_{n=0}^\infty\frac1{n!}\left( \tau \frac{\partial}{\partial t} \right)^n(gh) \\ &=&\sum_{n=0}^\infty\frac{\tau^n}{n!}\frac{\partial^n}{\partial t^n}(gh) \\ &=& \sum_{n=0}^\infty\frac{\tau^n}{n!}\sum_{k=0}^n\binom nk\frac{\partial^kg}{\partial t^k}\frac{\partial^{n-k}h}{\partial t^{n-k}} \\ &=& \sum_{n=0}^\infty\sum_{k=0}^n\frac{\tau^k}{k!}\frac{\partial^kg}{\partial t^k}\frac{\tau^{n-k}}{(n-k)!}\frac{\partial^{n-k}h}{\partial t^{n-k}} \\ &=& \sum_{k=0}^\infty\sum_{n=k}^\infty\frac{\tau^k}{k!}\frac{\partial^kg}{\partial t^k}\frac{\tau^{n-k}}{(n-k)!}\frac{\partial^{n-k}h}{\partial t^{n-k}} \\ &=& \sum_{k=0}^\infty\sum_{j=0}^\infty\frac{\tau^k}{k!}\frac{\partial^kg}{\partial t^k}\frac{\tau^j}{j!}\frac{\partial^jh}{\partial t^j} \\ &=& \sum_{k=0}^\infty\frac{\tau^k}{k!}\frac{\partial^kg}{\partial t^k}\sum_{j=0}^\infty\frac{\tau^j}{j!}\frac{\partial^jh}{\partial t^j} \\ &=& \left(\sum_{k=0}^\infty\frac{\tau^k}{k!}\frac{\partial^kg}{\partial t^k}\right)\left(\sum_{j=0}^\infty\frac{\tau^j}{j!}\frac{\partial^jh}{\partial t^j}\right) \\ &=& \left(\exp \left( \tau \frac{\partial}{\partial t} \right) g\right)\left(\exp \left( \tau \frac{\partial}{\partial t} \right) h\right)\;. \end{eqnarray*}