Is a language regular
(I think by pointers you mean the "pebbles" that are described in Kozen's Automata and Computability.)

The language $L'$ is regular.  

You can prove that as follows: Consider the DFA $D_1 = (Q, \delta, F, q_0)$ for $L$. We will create another automata $D_2 = (Q \times Q \times Q \times \{0,1\}, \Delta_2, F_2, I)$. 

$I = \{(q_0, q, q, 0) : q,q_0 \in Q\}$ 

$F = \{q, q_F, q, 0 : q_F \in F, q \in Q \}$

Let's first understand the intuition before defining the transition function $\Delta_2$:


We are constructing two layers: all the transitions in the first layer will be to some state in the second layer, and all those in the second layer will be toward the first layer. 
On reading an alphabet when on a state of the form $(q_1, q_2,q_2, 0)$, the state in the first entry will change as per the transition function of $D_1$. The state in the second entry will also follow $D_1$, but it will be symbol agnostic, in the sense that it can change to any state irrespective of the symbol. The third entry will remain the same. (It is used to save the state at which we started the second pebble). The fourth entry will change from 0 to 1 which is equivalent to moving from one layer to another.
When on a state of the form $(q_1, q_2,q, 1)$, all the transitions are $\epsilon$ transitions and move us back to $(q_1, q_2',q_2, 0)$ where $q_2'$ is a state which is reachable from the state $q_2$ in $D_1$. 


So in this automata, we first non-deterministically choose a start state of the form $(q_0, q, q,0)$. The state $q$ is the state from which we are reading $\beta$, and from $q_0$ we are reading the actual input $\alpha$. As the automata reads the word, for each letter of $\alpha$, we non-deterministically choose two letters for $\beta$ and run the same automata from that state in the second component. If on reading $\alpha$ we end up in the state we started reading $\beta$ (which is chosen non-deterministically), and the second component reaches a final state in $D_1$, then it means that the $\exists  \beta : \alpha \beta \in L$. And hence we accept.

Formally, 
\begin{equation}
\Delta_2((q_1, q_2, q, 0),a) = \{(\delta_1(q_1, a), \delta(q_2, x), q, 1) : x \in \Sigma)\}\\
\Delta_2((q_1, q_2, q, 1),\epsilon) = \{(q_1, \delta(q_2, x), q, 0) : x \in \Sigma)\}
\end{equation}. 



There is an easier way (as OP described):
Let $D_1'$ be the NFA obtained by reversing all the transitions in $D_1$ and allowing all symbols for each transitions (i.e., make it symbol agnostic). Take the states to be 3-tuples from $Q \times Q \times \{0,1\}$. Start from $(q_0, q_f, 0)$ where $q_f \in F$. On reading a letter, move from state $q_0$ as per $D_1$, and move two steps from $q_f$ according to the $D_1'$. The final states can be $(q,q,0)$ where $q\in Q$. 

You can try figuring out why this works.