My way of understanding it is this:
$\hat{\tau}(\lambda) = 1/\bar{X}$
$\operatorname{Var}(1/\bar{X}) = E((1/\bar{X})^2)-E(1/\bar{X})^2$
To find $E(1/\bar{X})$ we consider the mgf $E(\exp{\frac{t}{\bar{X}}})$ (I'm not sure here whether I can calculate the mgf of a function of r.v as $E(e^{tf(x)})$ )
as $T=\bar{X} \sim \operatorname{Pois}(n\lambda)$
$E(\exp{\frac{t}{\bar{X}}}) = e^{-n\lambda}\sum_{T=0}^\infty e^{t/T}(n\lambda)^T/T!$
This blows up as $e^{t/0}$ is $\infty$, thus the mean and variance are infinite.
The solution states the following:
Since $P(\bar{X} = 0) > 0$, we get that even the first moment is infinite (not to mention the second) and there is no finite variance.
I feel like it's talking about the same thing. But I wasn't sure whether I calculated the moment of a function of R.V correctly.
Your moment generating function for $f(\bar X)$, of $$\mathbb E\left[e^{tf(\bar X)}\right] = \sum\limits_x e^{tf(x)} \,\mathbb P(\bar X = x)$$ would be correct for a discrete distribution if you summed over all possible values of $\bar X$.
You have almost but not quite not quite done that, since $\bar X$ can take fractional values and does not have a Poisson distribution, though $n\bar X$ does have a Poisson distribution.
In this case it not really matter, since with $f(x)=\frac1x$ the sum fails at the start when $x=0$, as you get the term $e^{t/0}\mathbb \,P(\bar X = 0)$, i.e. an expression involving division by $0$, multiplied by a positive value, and you cannot handle this.