Let's prove that $a_3=a_1$. 
Note that the recurrence for $R$ translates to the following PDE for the generating $F(x,y):=\sum_{n,m\geq0} R(n,m,p,q) \frac{x^n}{n!}\frac{y^m}{m!}$:
$$\frac{\partial}{\partial x} F(x,y) = \frac{\partial}{\partial y} \bigg(F(x,y) + \frac{p}q(e^{qy} - 1)F(x,y)\bigg)$$
with the boundary condition $F(0,y) = e^y$. This PDE is well solvable in CAS like  Maple, it can be easily verified that $F(x,0)=\sum_{n\geq0} R(n,0,p,q) \frac{x^n}{n!}$ does indeed coincide with the derivative of the e.g.f for $a_1$ given in Ira's answer. QED

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Equality between $a_4$ and $a_5$ should probably follow from the [Salas and Sokal paper](https://arxiv.org/abs/2008.03070), but I'm too lazy to verify.

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**ADDED**. Here are the requested details of computation. It turns out that Maple is not so good at simplifying symbolic radicals (or I'm not familiar with the best practice here), I will use a combination of Maple and Sage:

    sage: S = maple('simplify(subs(y=0, rhs( pdsolve( [ diff(F(x,y),x)=diff(F(x,y),y) + p/q * diff( (exp(q*y)-1)*F(x,y),y), F(0,y) = exp(y) ], F(x,y) ) ) ))').sage()
    sage: p,q,x = S.variables()
    sage: ascii_art(S.simplify_real().canonicalize_radical().simplify_real())
                                     -1            
            q + 1                    ---           
            -----                     q            
              q   /     q*x      p*x\     x*(p + 1)
    (-p + q)     *\- p*e    + q*e   /   *e         
    -----------------------------------------------
                          q     p - 2*q            
                        ----- + -------            
                 q*x    p - q    p - q   p*x       
            - p*e    + q               *e          
    
    sage: ascii_art( diff(((p-q)/(p-q*exp(x*(p-q))))^(1/q),x).canonicalize_radical() )
                                     -1            
            q + 1                    ---           
            -----                     q            
              q   /     q*x      p*x\     x*(p + 1)
    (-p + q)     *\- p*e    + q*e   /   *e         
    -----------------------------------------------
                        q*x      p*x               
                   - p*e    + q*e                  

Sage slightly better in simplifying such expressions, and it left us just noticing that $\frac{q}{p-q}+\frac{p-2q}{p-q}=1$ to conclude equality of the two expressions.