Reduction from NP-complete problem to unknown complexity problem and vice-versa
I shall assume "$\le$" indicates a poly-time, many-one (i.e., Karp) reduction. This seems to be the case since you are referring to NP-completeness.


  If I show that $B \le A$ I can state that A is also NP-complete because the two required conditions are satisfied: (i) A is in NP (ii) i reduced a NP -complete problem to A.


If $B \le A$, that is, $B$ is reducible to $A$, then $A$ is necessarily $\mathbf{NP}$-hard, that is, any problem in $C \in \mathbf{NP}$ is reducible to $A$ (i.e., $C \le A$). This is because any problem $C \in \mathbf{NP}$ is reducible to $B$ (due to the $\mathbf{NP}$-completeness of $B$) and we have $C \le B \le A$; using the transitivity of (this form of) reduction we obtain $C \le A$. However, this does not at all imply that $A \in \mathbf{NP}$—otherwise, any $\mathbf{NP}$-hard problem would be $\mathbf{NP}$-complete! To prove $A$ is $\mathbf{NP}$-complete, thus, you would also have to prove $A \in \mathbf{NP}$ separately.


  If I show that $A \le B$ I can say that $B$ is at least hard as $A$, so $A$ is at least NP-complete, but can be harder. 


$A \le B$ means $A$ is reducible to $B$, which can be reformulated as $B$ being at least as hard as $A$, as you have stated. In no way does this imply that $A$ is $\mathbf{NP}$-complete. Granted, it could be the case, but, for all we know, $A$ could be in $\mathbf{P}$, a regular language, or even trivial (i.e., $\Sigma^\ast$ or $\varnothing$).