Is the NP-hardness Proof with One Way Implication Correct and Why?
Consider the following two wrong "reductions" from $SAT$ to $L = \{1\}$.  Let $\phi$ be a $SAT$ instance.

We "reduce" every $\phi$ to 1, i.e. $f(\phi) = 1$. Now we have $\phi \in SAT \Rightarrow 1 \in L$. But $L$ is clearly not NP-hard.

Next we "reduce" every $\phi$ to 0, i.e. $f(\phi) = 0$. Now we have $ 0\in L  \Rightarrow \phi \in SAT$. But again $L$ is clearly not NP-hard.

Thus we can see that for a correct reduction we need a proper if and only if condition: $x \in \Pi' \Leftrightarrow f(x) \in \Pi$.



If we have a reduction as above, and if we know that SAT is NP-complete (in other words a hard problem), we have two corollaries from this.

1) Any language can be reduced to SAT. So every language can be reduced to SAT and then to $L$, so $L$ is also NP-complete (now we know that $L$ must also be a hard problem).

2) Any algorithm for membership of $L$ can be used to solve membership to SAT. So if we have an "algorithm" for $L$ then so do we have an "algorithm" for SAT. The algorithm will work this way. From $\phi$ we will construct $f(\phi)$ and then apply our "algorithm" to $f(\phi)$. If answer is YES then $\phi$ is also satisfiable, and if answer is NO then $\phi$ is not satisfiable. 

So from 2) we can see that an algorithm for $L$ is applicable for $SAT$. But vice versa is not true. Any algorithm for $SAT$ (polynomial or exponential) can not be used for $L$ using this reduction (there might be other ways but not using just the above reduction). Notice the uncovered region in $L$. So this is what is meant by cryptic comment: solution of $L$ $\Rightarrow$ solution of SAT.