The incomplete beta function $B(x; a,b)$ is defined as
$$B(x;a,b) = \int_0^xt^{a-1}(1-t)^{b-1}dt.$$
Is there any identity which we can apply so that the difference $B(x+y;a,b)-B(x;a,b)$ can be simplified or expressed recursively in terms of (incomplete) beta functions of $x$ and $y$ only?
If we assume $a , b \in \mathbb{N}$ and introduce $c=a-1$ and $d=b-1$ (to make the final result fit in a single line), we can use the binomial theorem to find for every $x,y \in \mathbb{C}$ \begin{align} &\phantom{=} \mathrm{B} (x+y;c+1,d+1) - \mathrm{B} (x;c+1,d+1) = \int \limits_x^{x+y} t^{c} (1-t)^{d} \, \mathrm{d} t = \int \limits_0^{y} (s+x)^{c} (1-s -x)^{d} \, \mathrm{d} s \\ &= \sum \limits_{k=0}^{c} \sum \limits_{l=0}^{d} {c\choose k} {d\choose l} x^{c+d-k-l} (-1)^{d-l} \int \limits_0^{y} s^{k} (1-s)^{l} \, \mathrm{d} s \\ &= \sum \limits_{k=0}^{c} \sum \limits_{l=0}^{d} {c\choose k} {d\choose l} (-1)^{d-l} (c+d-k-l) \mathrm{B}(x; c+d-k-l,1) \mathrm{B}(y; k+1,l+1) \, , \end{align} setting $0 \cdot \mathrm{B}(x,0,1) = 1$ for consistency.
Using the binomial series, similar results can be derived for more general values of $a$ and $b$, but depending on the particular combination of $a$, $b$, $x$ and $y$ one has to deal with convergence issues and non-integer powers of complex numbers.
