The question requires interpretation. One reasonable thing is to suppose that the bird laying the eggs is of Species $1$ with probability $p$, and of Species $2$ with probability $1-p$. Perhaps, quite unreasonably, you are expected to assume that $p=\frac{1}{2}$.
Then if $X$ is the number of eggs, $$\Pr(X=k)=pe^{-\lambda_1}\frac{\lambda_1^k}{k!}+(1-p)e^{-\lambda_2}\frac{\lambda_2^k}{k!}.$$ If $\lambda_1\
e \lambda_2$, this is not Poisson for any "mixed" $\lambda$, except in the extreme cases $p=0$ or $p=1$.
**Remark:** I am not sure what you meant by the sum of the Poisson functions. Maybe you meant $\Pr(X=k)=e^{-\lambda_1}\frac{\lambda_1^k}{k!}+e^{-\lambda_2}$. If so, that is not quite right, though if we take $p=\frac{1}{2}$, as you may be intended to do, it is very close. But we need to divide by $2$. In general $\Pr(X=k)$ is a weighted average of the two mass functions.