Are different assignments allowed for the implication graph proof of 2-SAT being in P?
You have to understand that $a \implies b$ cannot be interpreted as $\neg a \implies \neg b$. Only the contraposition $b \implies a$ is true, but this one is already in the implication graph by construction.
Thus, with your algorithm, you will make abrirtrary assignments. This will probably leads to an impossibility even if the problem does have a solution. Of course, on trivial examples, one can see that this algorithm may produce a solution.
Also note that  $a \implies b \implies c$ means there is elsewhere in the graph $\neg c \implies \neg b \implies \neg a$. So when you decide to assign false in order to $a, b, c$, you are reading the graph and the strongly connected components reversly.