I know that for binomial and negative binomial RVs there is an additive property where
if $X_1\sim bin(a, p)$ and $X_2\sim bin(b, p)$ then $X_1+X2 \sim bin(a+b, p)$
if $Y_1\sim NB(c, p)$ and $Y_1\sim NB(d, p)$ then $X_1+X2 \sim NB(c+d, p)$
Is there a similar additive property for hypergeometric RVs?
In the binomial and negative binomial examples, you missed the important condition of the independence of $X_1$ and $X_2$.
Anyway, the hypergeometric would not possess this property in any obvious way, because if you had two sets of $N_1$ and $N_2$ objects among which respectively $M_1$ and $M_2$ are of one type (and hence respectively $N_1-M_1$ and $N_2-M_2$ are of another), and you independently drew SRSWOR samples of sizes $n_1$ and $n_2$ from the two sets respectively, the total number of objects of the first type would not have a $\mathrm{HG}(N_1+N_2,M_1+M_2,n_1+n_2)$ distribution: the latter distribution would arise if the two sets were combined, and from the combined set a WOR sample of size $n_1+n_2$ were drawn.
For a simple example, suppose $N_1=N_2=3$, $M_1=M_2=2$ and $n_1=n_2=1$: $X_1$ and $X_2$ represent the numbers of white balls, say, when one ball each is drawn from two urns each containing two white and one black balls. Then the probability that $\{X_1+X_2=2\}$ is $P(X_1=1,X_2=1)=\left(\frac 23\right)^2$ whereas the probability that a $\mathrm{HG}(6,4,2)$ variable takes the value 2 is $\frac{{4\choose 2}}{{6\choose 2}}$, a different value.
