"Regular languages over a common alphabet are closed under union."
I see two possible points of confusion in your question, and I will address them separately.


What is meant by the title of your post: ""Regular languages over a common alphabet are closed under union."
"The union of $L_1,L_2$ is $\{x:x∈L_1∨x∈L_2\}$ 
Does this mean that, for any string $s∈L_1$, we also have $s∈L_2$?"


What is "Closure Under Union"?

Regular languages are a class of languages that exhibit the closure property with the union operation.

The closure property means that when an operation is applied to language(s) of a certain class the resulting language will also be of that class.

In this case the class of languages is the regular languages and the operation is union.

So all this means is that the union of two regular languages is also a regular language.

What is the union of two languages?

Languages can be thought of as sets of strings, so the union operation works the same way as in typical Set Theory.

I would read $\{x:x∈L_1∨x∈L_2\}$ as "the set of strings such that the string is in $L_1$ or the string is in $L_2$."

This does NOT mean that for any string $s∈L_1$, we also have $s∈L_2$.

We only have that for any string $s$ in the union of languages $L_1$ and $L_2$, denoted $s \in (L_1\cup L_2$), that string $s$ is in either $L_1$ or $L_2$.