Check if $L$ is regular then $L'$ is regular (and and vice versa) let's define $w\leftrightarrow v$. It does mean that  we may  "create" $w$ using word $v$ in following way:
every single letter $x\in \Sigma$ in word $v$ may be replaced by $xx$, and every double letter $xx$ in word $v$ may be replaced by single $x$.
example: $aabbbaaabba\leftrightarrow ababa$
$abbaa\leftrightarrow aaba$
$L'=\{w|w\leftrightarrow v \text{ for some $v\in L$}\}$  
And now, check if is true:
(1) $L\in Reg \rightarrow L'\in Reg$ 
I can show it by construcntion. I consider if exist other way ? Maybe Myhill-Nerode ?
(2) $L'\in Reg \rightarrow L\in Reg$
My Idea
Let's consider equivalent implicaton:  $L\notin Reg \rightarrow L'\notin Reg$
It isn't true, due to: $n\ge 0,\ \ L=0^n1^n\notin Reg$; Then $L'=0^*1^*\in Reg$.
So due to the fact that equivalent implications is not true,  $L'\in Reg \rightarrow L\in Reg$   is not true.
I shown it because I'm not sure the correctness of this.
(3) $L\in CFG \rightarrow L'\in CFG$
What about it ?
 A: First of all, let me state your question a little more precisely. Let $A$ be a finite alphabet and let $\sim$ be the congruence on $A^*$ generated by the relations $a \sim a^2$, for each letter $a$. Then if $u = a_1^{k_1}a_2^{k_2} \dotsm a_n^{k_n}$, where $k_1, ..., k_n > 0$ and two consecutive letters $a_i, a_{i+1}$ are always distinct, then 
$$
[u]_\sim = \{a_1^{r_1}a_2^{r_2} \dotsm a_n^{r_n} \mid r_1, ..., r_n > 0\}
$$
Given a language $L$ of $A^*$, let
$$
[L]_\sim = \{ u \in A^* \mid \text{there exists $v \in L$ such that $u \sim v$}\}.
$$
Let $\mathcal{P}(A^*)$ be the monoid of all languages on $A^*$ under (concatenation) product. Let $\sigma:A^* \to \mathcal{P}(A^*)$ be the monoid morphism defined, for each letter $a \in A$, by $\sigma(a) = a^+$. Thus, for instance, $\sigma(aabacb) = a^+a^+b^+a^+c^+b^+$. Setting
$$
\sigma^{-1}(L) = \{ u \in A^* \mid \sigma(u) \cap L \not= \emptyset\}
$$
I claim that
$$
[L]_\sim = \sigma(\sigma^{-1}(L))
$$
The inclusion $\sigma(\sigma^{-1}(L)) \subset [L]_\sim$ is obvious. For the opposite inclusion, let $v \in [L]_\sim$. Then there exists a word  $u$ of $L$ such that $u \sim v$. Thus $v  = a_1^{r_1}a_2^{r_2} \dotsm a_n^{r_n}$ and $u = a_1^{k_1}a_2^{k_2} \dotsm a_n^{k_n}$ for some $r_1, ..., r_n > 0$ and $k_1, ..., k_n > 0$. Now $a_1a_2  \dotsm a_n \in \sigma^{-1}(L)$ since the language $\sigma(a_1a_2  \dotsm a_n) = a_1^+a_2^+  \dotsm a_n^+$ contains $u$. Moreover, $v \in \sigma(a_1a_2  \dotsm a_n)$ and thus $v \in \sigma(\sigma^{-1}(L))$.
Now, you can observe that both $\sigma$ and $\sigma^{-1}$ are rational transductions and rational transductions preserve regular languages and context-free languages. It follows that if $L$ is regular, then $[L]_\sim$ is regular and if
$L$ is context-free, then $[L]_\sim$ is context-free. 
