Maybe this is too obvious to be interesting, but if we have bounds $\frac{1}{n!} \left| \frac{\partial^n}{(\partial x)^n} f(x,y) \right| < C_n$ on all of $(a,b) \times (c,d)$ such that $\sum C_n r^n$ converges for some $r>0$, then we get the conclusion.
This is because the Taylor series with respect to $x$ then converges on circles of radius $r$ around any point of $(a,b)$. Write $U$ for the open subset of points in $\mathbb{C}$ which are within distance $r$ of some point of $(a,b)$. These Taylor series give an extension of $f$ to $U \times (c,d)$, holomorphic in the first variable. (They agree on overlaps, since they agree on the real interval where they overlap.) Moreover, $|f| \leq \sum C_n r^n$ everywhere on $U \times (c,d)$.
We want to claim that $\int_c^d f(x,y) dy$ is holmorphic on $U$. By Morera's theorem, it is enough to check that $\int_{x \in \gamma} \int_{y=c}^d f(x,y) dy dx=0$ for any closed curve $\gamma \subset U$. By Fubini, we may switch the integral to $\int_{y=c}^d \int_{x \in \gamma} f(x,y) dx dy$, and the inner integral is zero since $f$ is holomorphic. (We may apply Fubini because we have a uniform bound for $f$ throughout $U \times (c,d)$.)