The questions are a bit unclear. I think using power analysis and standard hypothesis testing should be sufficient to answer the question. Let me start by rephrasing the first questions:
- "How many units do we need to see produced in minimum before we can be confident (at p=0.05) in saying that the process changes were effective in reducing defects?"
To answer the question, we would like to construct a test with the null hypothesis: "The process changes were not effective, i.e. $\lambda_0=1/1000$." We would be confident to reject the null when the observation i.e. n observations without defective is an extreme event under the null hypothesis ($p\leq 0.05$). Notice that in order to keep the observed units at minimum to test the hypothesis, I would not want to see any defectives during the inspection. However, of course, you can observe some defective during the inspection and you can still test the null hypothesis, the sample size would increase with the number of defectives you allow to observe.
Under the null hypothesis $\lambda_0=1/1000$, the probability of observing 0 defectives within an interval $\Delta t$ is
$$P(N(t+\Delta t)-N(t))=e^{-\lambda_0\Delta t}$$
You can solve for the $\Delta t$ letting $p\leq 0.05$, which is $\Delta t^* \geq 2996$. Phrased differently, observing 2996 units without any defectives is an extreme event under the defective rate of 1/1000 such that that the defective rate is likely to be smaller than 1/1000 (one-tail test).
Notice that I did not claim that the test result indicates $\lambda=1/3000$, and we could not make such assertion under current test setting. But we can compute the power of the test, which is the probability of observing no defectives for 2996 units (reject the null hypothesis) when the alternative i.e. $\lambda=1/3000$ is true.
$$P(N(t+\Delta t^*)-N(t)=0|\lambda_1 =\frac{1}{3000})=e^{-\lambda_1\Delta t^*}$$
According to the power computation formula, the power decreases with the number of units. The rejection space $\Delta t^*=2996$ result in power $\approx 0.37$, i.e. there is 0.63 chance that you will not reject the null when the alternative is true.
To increase the power of the test, you can increase the type 1 error ($\alpha$ and $\beta$ trade-off). For a power of 0.8, you need 670 units observed without defectives (but your type 1 error claiming there is reduction in defective rate is 51%).
