Can two claims be Logically equivalent and have different truth tables? I'm looking for an example that will help me understand better the issue of logical equivalence.
I know that two claims can be logically equivalent and yet have different truth tables.
But while trying to think of an example for this a few questions popped.
What does it mean for two truth tables to be distinct (unequal)?
I think $P \to Q \equiv \lnot P \lor Q$ is a good example of claims which are equivalent, but don't have equivalent truth-tables. 
But what about $P \lor Q$ and $Q \lor P$?
Is there a better way to understand this?
 A: NO. If two statements are logically equivalent, then their columns in the combined truth-table will be exactly the same. Truth-tables are a way of showing the truth-conditions of statements, and two statements are equivalent if and only if they have the same truth-conditions ... i.e. have the same truth-table!
By 'combined truth-table I mean that both statements would be put in the same truth-table. Because if you don't do that, then superficially it may look like the truth-tables of, for example, $P$ and $P \land (P \lor Q)$ are not the same:
$\begin{array}{c|c}
P&P\\
\hline
T&T\\
F&F\\
\end{array}$
$\begin{array}{cc|c}
P&Q&P\land (P \lor Q)\\
\hline
T&T&T\\
T&F&T\\
F&T&F\\
F&F&F\\
\end{array}$
But using a combined truth-table:
$\begin{array}{cc|c|c}
P&Q&P&P \land (P \lor Q)\\
\hline
T&T&T&T\\
T&F&T&T\\
F&T&F&F\\
F&F&F&F\\
\end{array}$
A: No.  In fact, what is true is that  two claims have equivalent truth-tables if and only if they are equivalent. And two statements are not logically equivalent if and only if they do not have equivalent truth-tables.
You claim that $P\to Q \equiv \lnot P \lor Q$ is an example in which the two statements are logically equivalent, but do not  have equivalent truth-tables. Let's explore that:
\begin{array}{cc|c}
P&Q|&P\to Q\\
\hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\
\end{array}
\begin{array}{cc|c}
P&Q&\lnot P \lor Q\\
\hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\
\end{array}
Of course, we could have written $\lnot P \lor Q$ using the following truth-table:
\begin{array}{cc|c|c}
P&Q&\lnot P& \lnot P \lor Q\\
\hline
T&T&F&T\\
T&F&F&F\\
F&T&F&T\\
F&F&T&T\\
\end{array}
While we used the extra column for the last claim, we could put $\lnot P$ as a separate column in the first claim, e.g.:
\begin{array}{\color{blue}{cc}|c|c}
P&Q&\lnot P&P\to Q\\
\hline
T&T&F&T\\
T&F&F&F\\
F&T&T&T\\
F&F&T&F\\
\end{array}
In any case the two truth-tables are equivalent when under the same truth-value assignment of variables, the right-most columns (under each claim) are identical.

$P\lor Q \equiv Q\lor P$, and we know that for the simple fact that $\lor$ is commutative.  
And indeed, their truth tables also reveals they are equivalent (i.e., proves the top statement), meaning that the right-most columns (with the header $P\lor Q$, $Q\lor P$, respectively), evaluate to the same truth value under the same truth value assignments to the variables $P, Q$.
E.g
\begin{array}{cc|c}
P&Q&P\lor Q\\
\hline
T&T&T\\
T&F&T\\
F&T&T\\
F&F&F\\
\end{array}
\begin{array}{cc|c}
P&Q&Q\lor P\\
\hline
T&T&T\\
T&F&T\\
F&T&T\\
F&F&F\\
\end{array}
