From a theoretical point of view it is exactly the distribution of $X_1 + X_2.$ But it is not binomial because the two Success probabilities are not the same.
$$E(X_1 + X_2) = n_1p_1 + n_2p_2 = 50 + 60.$$
Provided $X_1$ and $X_2$ are independent, you also have
$$Var(X_1 + X_2) = n_1p_1(1-p_1) + n_2p_2(1-p_2).$$
[In particular, this is not the same thing as $(n_1 + n_2)p_a(1-p_a),$ where
$p_a = (p_1 + p_2)/2.$]
If this were an applied situation and the 100 observations for $X_1$ were
taken contemporaneously with the 100 observations for $X_2$ and in the same place, then I'd investigate circumstances to make sure about the independence.
For example, if the subjects for $X_1$ are 100 randomly chosen men from a population in which 50% are Democrats, and the subjects for $X_2$ are 100 randomly chosen women from a population in which 60% are Democrats, then independence seems reasonable. But I'd want to make sure that 'for
convenience' the researchers didn't select 100 married M/F couples and
use the 100 men and 100 women from those.