= Solution
Given the complete stopped sample $(x_1,\ldots,x_n)$, the <likelihood function> is
$$
L(\theta;\mathbf x)=\theta^ne^{-\theta S}\alpha(x_n)\prod_{i=1}^{n-1}[1-\alpha(x_i)],\qquad S=\sum_{i=1}^nx_i.
$$
The stopping factor is positive and independent of $\theta$. The <Fisher-Neyman factorization theorem> makes $(N,S)$ sufficient. For two possible stopped samples of lengths $n,m$ and sums $S,S'$, their likelihood ratio is a parameter-free factor times $\theta^{n-m}e^{-\theta(S-S')}$. This is independent of $\theta$ exactly when $n=m$ and $S=S'$: differentiating its logarithm gives $(n-m)/\theta-(S-S')=0$ for all $\theta>0$. By the likelihood-ratio criterion,
$$
\boxed{(N,\sum_{i=1}^NX_i)\text{ is a minimal sufficient statistic}.}
$$
The <likelihood principle> says proportional likelihoods give the same evidence about the parameter. Thus with the full sample observed, this parameter-free stopping rule adds no likelihood evidence beyond the realized values and length; it gives the same likelihood-based inference as a fixed-length sample with those values. This statement does not assert that frequentist sampling distributions are unchanged.
Let $q(\theta)=\int_0^\infty\alpha(x)\theta e^{-\theta x}\,dx\in(0,1)$ be the chance of stopping at a given stage. Independence of successive stages gives
$$
p_Y(y\mid\theta)=\sum_{n\geq1}(1-q(\theta))^{n-1}\alpha(y)\theta e^{-\theta y}=\frac{\alpha(y)e^{-\theta y}}{\int_0^\infty\alpha(x)e^{-\theta x}\,dx}.
$$
Therefore $a(y)=\log\alpha(y)$ and $k(\theta)=\log\int_0^\infty e^{a(y)-\theta y}\,dy$ give the requested <exponential family> form.
Writing $p=p_Y$, differentiation in $y$ gives $p'=(a'-\theta)p$. Integrating and using $p(0)=p(\infty)=0$ shows $\mathbb E_\theta a'(Y)=\theta$. Likewise
$$
p''=[a''+(a'-\theta)^2]p.
$$
The stipulated vanishing of $p'$ at both boundaries gives
$$
\boxed{\operatorname{Var}_\theta(a'(Y))=-\mathbb E_\theta a''(Y).}
$$
These integrations require the displayed moments and integrals to exist, the usual regularity interpretation of the question; when establishing a finite variance this assumption should be explicit.
Differentiating the normalization in $\theta$ gives $k'(\theta)=-\mathbb E_\theta Y$ and $k''(\theta)=\operatorname{Var}_\theta Y$. The <score function> is $-k'(\theta)-Y$, so the <Fisher information> is $I(\theta)=k''(\theta)$. The <Cramér-Rao bound> states that a regular unbiased estimator of $\theta$ has variance at least $1/I(\theta)$. Apply it to $a'(Y)$:
$$
\boxed{-k''(\theta)\,\mathbb E_\theta a''(Y)\geq1.}
$$
Equivalently differentiate $\mathbb E_\theta a'(Y)=\theta$ and use the <Cauchy-Schwarz inequality> to get $1\leq\sqrt{\operatorname{Var}(a'(Y))\operatorname{Var}(Y)}$, which gives the same conclusion.
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