I have noted that when $\lvert A\cap C\rvert = \lvert B\cap C \lvert = n$ for a finite $n$ number, the proposition holds, but if the intersection is infinite, I am not sure.
I know that there are three bijections by hypothesis, $f: A \rightarrow B$. $g: C \rightarrow D$ and $h: A \cap C \rightarrow B \cap D$
I tried defining a function $L: A \cup C \rightarrow B \cup D$ as
$ L(x) =\begin{cases} f(x), & \text{if $x$ $\in A \setminus C $} \\ h(x), & \text{if $x$ $\in A \cap C$} \\ g(x), & \text{if $x$ $\in C \setminus A $} \end{cases} $
so that the first case would go to $B\setminus D$, the second to $B \cap D$ and the third one to $D\setminus B$ and as a result, $L$ would have image $B \cup D = (B\setminus D) \cup (B \cap D)\cup (D\setminus B) $. But then I realized I don't know where $f$, $g$ or $h$ are sending the elements I'm taking in $A \cup C$, they could be sending them to the three disjoint sets I mentioned respectively , but that is probably not the case.