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
Equality between $a_4$ and $a_5$ should probably follow from the Salas and Sokal paper, but I'm too lazy to verify.
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.