= Solution
For a target $a(\theta)$, the <bias> of an integrable <estimator> $T$ is $\mathbb E_\theta T-a(\theta)$; it is an <unbiased estimator> if this is zero at every allowed parameter value. Here $Y$ has a <zero-truncated Poisson distribution>, $\theta>0$, and $p=1-e^{-\theta}\in(0,1)$. Write an estimator based only on $Y$ as $t(Y)$, assuming a finite expectation for every $\theta>0$. Unbiasedness requires
$$
\sum_{y=1}^\infty t(y)\frac{\theta^y}{y!}=(e^\theta-1)(1-e^{-\theta})=e^\theta+e^{-\theta}-2.
$$
Absolute integrability at every positive parameter ensures that the power series on the left converges absolutely on every complex disc. Uniqueness of coefficients in a <power series> therefore gives $t(y)=0$ for odd $y$ and $t(y)=2$ for positive even $y$. Conversely these values give the displayed identity, so
$$
\boxed{T=2\mathbf1_{\{Y\text{ even}\}}}
$$
is the unique <unbiased estimator> of this form. The uniqueness claim concerns nonrandomized functions of $Y$; allowing external randomness would permit addition of independent mean-zero noise.
For this <parity estimator for a zero-truncated Poisson count>, usefulness depends on the loss, but under <squared-error loss> it performs poorly despite being an <unbiased estimator>. Since $\mathbb P(Y\text{ even})=p/2$, $\mathbb E T^2=2p$ and
$$
\operatorname{Var}T=2p-p^2=1-e^{-2\theta}.
$$
It takes the inadmissible value $2$ with positive probability, and its <variance> tends to one rather than zero as $\theta\to\infty$. Clipping to the parameter interval gives $T_c=\min(T,1)=\mathbf1_{\{Y\text{ even}\}}$. This has <bias> $-p/2$, but
$$
\mathbb E[(T_c-p)^2]=p/2<2p-p^2=\mathbb E[(T-p)^2],\qquad0<p<1.
$$
Thus a simple biased <estimator> strictly improves its <mean squared error> at every parameter value; uniqueness among unbiased estimators does not make it optimal for this loss.
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