OR to XOR algebraically
The OR operation is associative:

$$ (a \vee b) \vee c $$

Apply your rule to \$(a \vee b)\$:

$$ (a \oplus b \oplus ab) \vee c $$

We can then apply your rule again, but one of the terms is the expression \$ (a \oplus b \oplus ab) \$:

$$ (a \oplus b \oplus ab) \oplus c \oplus (a \oplus b \oplus ab)c $$

Now it's just a matter of simplification. AND distributes over XOR, so we can distribute the \$c\$ in the final \$ (a \oplus b \oplus ab)c \$ and get:

$$ (a \oplus b \oplus ab) \oplus c \oplus (ac \oplus bc \oplus abc) $$

Because XOR is also associative we can drop all the parenthesis and reorder however we want:

$$ a \vee b = a \oplus b \oplus c \oplus ab \oplus ac \oplus bc \oplus abc$$

As Majenko's answer explains, this is every possible combination of one or more of the terms. This pattern extends to as many terms as you want, although it gets very long, very fast. If we want to tack on a \$\vee d\$, then following the same steps, we get:

$$ (a \oplus b \oplus c \oplus ab \oplus ac \oplus bc \oplus abc) \vee d \\
(a \oplus b \oplus c \oplus ab \oplus ac \oplus bc \oplus abc) \oplus d
  \oplus (a \oplus b \oplus c \oplus ab \oplus ac \oplus bc \oplus abc)d \\
(a \oplus b \oplus c \oplus ab \oplus ac \oplus bc \oplus abc) \oplus d
  \oplus (ad \oplus bd \oplus cd \oplus abd \oplus acd \oplus bcd \oplus abcd) \\
a \oplus b \oplus c \oplus d
  \oplus ab \oplus ac \oplus ad
  \oplus bc \oplus bd
  \oplus cd
  \oplus abc \oplus abd \oplus acd \oplus bcd
  \oplus abcd
$$

You can keep doing this, but at some point you won't be able to cross your eyes far enough.