Explicitly, let $X\sim \textrm{Pois}(\lambda)$ so that
$$P(X=n) = \frac{\lambda^ne^{-\lambda}}{n!}, \quad n\in \mathbb{Z}_{\ge0}, \ \lambda >0$$
and let $Y|X \sim \textrm{BetaBin}(n, \alpha, \beta)$ with
$$ P(Y=k | X=n) = \binom{n}{k}\frac{B(k+\alpha, n-k+\beta)}{B(\alpha, \beta)}, \quad n\in \mathbb{N}, \ \alpha, \beta > 0 $$
where $B(a, b) = \int_0^1 x^{a-1} (1-x)^{b-1} dx$ is the beta function. Then the product of these two is
$$ P(Y=k | X = n)\cdot P(X=n) = \frac{\lambda^ne^{-\lambda}}{n!} \binom{n}{k}\frac{B(k+\alpha, n-k+\beta)}{B(\alpha, \beta)}.$$
Using the total law of probability, the marginal distribution is given by $$P(Y=k) = \sum_{n=k}^\infty \frac{\lambda^n e^{-\lambda}}{n!}\binom{n}{k}\frac{B(k + \alpha, n -k+\beta)}{B(\alpha, \beta)}.$$
Is there a well-known, closed formula for this distribution?
Here's my attempt. We factor out the terms not dependent on $n$ and write out the binomial coefficient $$ P(Y=k) = \frac{e^{-\lambda}}{B(\alpha, \beta)} \sum_{n=k}^\infty \frac{\lambda^n}{n!}\frac{n!}{k! (n-k)!}B(k + \alpha, n -k+\beta).$$ Re-index $i = n - k$ $$ P(Y=k) = \underbrace{\frac{\lambda^k e^{-\lambda}}{k!}}_{=P(X=k)}\cdot\frac{1}{B(\alpha, \beta)} \sum_{i=0}^\infty \frac{\lambda^i}{i!}B(k + \alpha, i+\beta)$$ Insert $B(k+\alpha, \beta)$ in the numerator and denominator $$ P(Y=k) = P(X=k)\cdot \frac{B(k+\alpha, \beta)}{B(\alpha, \beta)} \sum_{i=0}^\infty \frac{\lambda^i}{i!}\frac{B(k+\alpha, i+\beta)}{B(k+\alpha, \beta)}.$$ The beta function in can be expressed in terms of gamma functions and then Pochhammer symbols using the identities $$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, \quad (x)_a = \frac{\Gamma(x+a)}{\Gamma(x)}$$ so that
$$\frac{B(k+\alpha, \beta)}{B(\alpha, \beta)} = \frac{(\alpha)_k}{(\alpha+\beta)_k}, \quad \frac{B(k+\alpha, i+\beta)}{B(k+\alpha, \beta)} = \frac{(\beta)_i}{(k+\alpha+\beta)_i}$$ Plugging these in, we have $$P(Y=k) = P(X=k)\cdot \frac{(\alpha)_k}{(\alpha+\beta)_k}\sum_{i=0}^\infty \frac{(\beta)_i}{(k+\alpha+\beta)_i}\cdot \frac{\lambda^i}{i!}.$$
The series is a confluent hypergeometric function of the first kind which has a couple notations
$$\sum_{i=0}^\infty \frac{(\beta)_i}{(k+\alpha+\beta)_i}\cdot \frac{\lambda^i}{i!} = {}_1F_1(\beta, k+\alpha+\beta, \lambda) = M(\beta, k+\alpha+\beta, \lambda).$$ This is as simplified as I could make it $$P(Y=k) = P(X=k) \cdot \frac{(\alpha)_k}{(\alpha+\beta)_k}\cdot M(\beta, k+\alpha+\beta, \lambda).$$
Numerically, this is likely a probability mass function. Does this simplify further, or are there recursive identities for the confluent hypergeometric function that make can be expressed in terms of more well-known distributions? My guess is there should be some convolution approach because the Poisson distribution factored out separately.