It isn't a bivariate Poisson precisely, but has a pmf of:
$$ \mathbb{P}(X=x, Y=y)=\frac{\lambda^y e^{-2 \lambda}}{x !(y-x) !} I_{\{0, \ldots, y\}}(x) I_{\mathbb{N}}(y), \lambda>0$$
When I tried to evaluate $\mathbb{P}(Y=y) $ I ended up reaching a sum I don't know how to handle:
$$ \begin{aligned} P(y=y) & =\sum_{i=0}^{\infty} P(x=i, y=y) \\ & =\sum_{i=0}^{\infty} \frac{\lambda^y e^{-2 \lambda}}{i !(y-i) !} I_{\{0, \ldots, y\}}(i) I_{\mathbb{N}}(y) \\ & =x^y e^{-2 \lambda} I_{\mathbb{N}}(y) \sum_{i=0}^{\infty} \frac{1}{i !(y-i) !} I_{\{0, \ldots, y\}}(i) \\ & =\lambda^y e^{-2 \lambda} I_{\mathbb{N}}(y) \sum_{i=0}^y \frac{1}{i !(y-i) !} \end{aligned} $$
Am I going in the right direction? Or is there another argument I could be thinking about?
You almost there. Here is how to calculate the sum: \begin{align*} \sum_{i = 0}^y \dfrac{1}{i!(y - i)!} &= \dfrac{1}{y!}\sum_{i = 0}^y \dfrac{y!}{i!(y-i)!}\\ &= \dfrac{1}{y!}\sum_{i = 0}^y C^i_y \end{align*} Recall the Newton's binomial theorem, $$ (a + b)^y = \sum_{i = 0}^y C^i_y a^i b^{y - i} $$ By choosing $a = b = 1$, we have $$ \sum_{i = 0}^y C_y^i = 2^y $$ Thus, $$ \mathbb{P}(Y = y) = e^{-2\lambda} \dfrac{(2\lambda)^y}{y!}1_{\mathbb{N}}(y) $$ or, $Y \sim \mathcal{P}(2\lambda)$