Can a RE language be reduced to a non-RE language?
First, note that if $p$ and $q$ are any two statements, the implication $p\implies q$ is logically equivalent to $(\lnot q)\implies(\lnot p)$. This equivalence is known as "taking the contrapositive".

Because of this, your statements $(I)$ and $(II)$ are equivalent. That is, the first two properties you mention are equivalent (and they follow from $L \leq_m L'$). The second two properties are also equivalent. Your question then becomes:

$(I)$ If $L \leq_m L'$ and $L \in RE$, can we conclude $L' \not\in RE$ ?

In general, no, we can not conclude that.

For instance, take $L=\{0\}$. This is a decidable language (we can test the input string against $0$, and accept iff they are equal), so it is also a $RE$ language.

Now, note that we can many-one reduce $L$ to absolutely any set $L'$ except the empty set and the full set ($\Sigma^*$). This includes non-$RE$ sets like the complement of the halting problem.

Indeed, to find a reduction, take any word $w\in L'$ and any other word $x\notin L'$. Then, the reduction function behaves as follows: it tests its input, if it is equal to $0$, then it returns $w$, otherwise it return $x$.

By construction, this is a reduction, so $L \leq_m L'$, even if $L\in RE$ and $L'\notin RE$.