First, you need to understand the definition like this. Pool all the propositional letters (sentence symbols) which occur in wffs in the set $S$ together with the (same or different!) propositional letters which occur in $\tau$. So if $S$ contains the wffs $P$ and $(\neg P \lor Q)$, and $\tau$ is $(Q \lor R)$, then the relevant letters that occur somewhere or other are $P, Q, R$. Now consider valuations $V$ of these letters. $S \vdash \tau$ iff for every such $V$, if $V$ makes all the wffs in $S$ true, it makes $\tau$ true.
So, for example, we indeed have
$$\{P, (\neg P \lor Q)\} \vDash (Q \lor R),$$
even though the conclusion contains a propositional letter which isn't in the premisses. (And that's what we want! For the conclusion plainly does logically follow from the premisses!)
Second, suppose though that none of the propositional letters in $S$ are in $\tau$. To take a simple case, suppose $S$ contains the wff $P$, and $\tau$ is $Q$. Then we would not have
$$\{P\} \vDash Q$$
because the valuation $P \Rightarrow T$, $Q \Rightarrow F$ makes the sole premiss true and the supposed conclusion false, showing this isn't a tautological entailment.
In fact we'll have the following result. If $S$ and $\tau$ lack propositional letters in common,
$$S \vDash \tau$$
holds just when $\tau$ is already a tautology, or when the wffs $S$ are contradictory (cannot all be true together), but not otherwise.
You have to check the definition of tautology.
Usually, tautology is defined in the context of propositional logic.
For first order logic, a formula is a tautology if it is a formula obtainable from a tautology of propositional logic by replacing (uniformly) each sentence symbol by a formula of the first-order language.
Thus, $\forall x P(x) \to \forall x P(x)$ is a tautology, being an "instance" of $A \to A$, while $x=x$ is not, because it is an "instance" of the single sentence symbol $A$, which is not a tautology.
Of course, $x=x$ is valid, i.e. true in every interpretation.
Best Answer
'Tautological' means 'in virtue of its truth-functional properties'.
$a=a$ is a perfectly good example. It is a logically necessary truth, but it is not true on the basis of any truth-functional operators involved. So, it is not a tautologically necessary truth.
This is why first-order logic (FOL) is more powerful than propositional logic (PL): FOL can 'see' things that PL cannot.
OK, one more example: 'Either everything is purple, or there is something that is not purple'. This is not a tautological truth: again, PL is unable to 'understand' the quantifiers involved here that make this a logically necessary truth.