I only found for: m = n = a = b = 0,\[Rho] < 1

f[a_, b_, \[Rho]_, m_, n_] := NIntegrate[(E^((-x^2 - y^2 + 2 x y \[Rho])/(2 (1 - \[Rho]^2))) x^m y^n)/(
2 \[Pi] Sqrt[1 - \[Rho]^2]), {x, a, Infinity}, {y, b, Infinity}]
f[0, 0, 1/2, 0, 0]
(*0.333333*)
(\[Pi] + 2 ArcSin[\[Rho]])/(4 \[Pi]) /. \[Rho] -> 1/2 // N
(*0.333333*)

I think that does not have finite closed-form expression in terms of very large class of special functions.

If a=b=0,m>0,n>0 and \[Rho] < 1 then:

 g[m_, n_, \[Rho]_] := 
  1/\[Pi] 2^(1/2 (-6 + m + n)) (1 - \[Rho]^2)^(
   1/2 (1 + m + 
      n)) (2 Gamma[(1 + m)/2] Gamma[(1 + n)/2] Hypergeometric2F1[(
       1 + m)/2, (1 + n)/2, 1/2, \[Rho]^2] - (
     m n Gamma[m/2] Gamma[n/
       2] (-(-1 + \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/
          2, -(1/2), \[Rho]^2] + (-1 + (4 + m + 
              n) \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/2, 1/
          2, \[Rho]^2]))/((1 + m) (1 + n) \[Rho]))
g[2, 3, 1/2] // N
(* 1.34643*)
f[0, 0, 1/2, 2, 3]
(* 1.34643*)

Regards M.I.

PS. I have another integral:

Integrate[(
  E^((-x^2 - y^2 + 2 x y \[Rho])/(2 (1 - \[Rho]^2))) x^m y^n)/(
  2 \[Pi] Sqrt[1 - \[Rho]^2]), {x, -Infinity, 
   Infinity}, {y, -Infinity, Infinity}] == 
 1/\[Pi] 2^(1/2 (-4 + m + n)) (1 - \[Rho]^2)^(
  1/2 (1 + m + n)) ((
    n Gamma[1 + m/2] Gamma[n/
      2] (-(-1 + \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/
         2, -(1/2), \[Rho]^2] + (-1 + (4 + m + 
             n) \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/2, 1/
         2, \[Rho]^2]) (-1 + Cos[m \[Pi]] + I Sin[m \[Pi]]))/((1 + 
       m) (1 + n) \[Rho]) + 
    1/((1 + m) (1 + n) \[Rho]) (-1)^(1 + n)
      n Gamma[1 + m/2] Gamma[n/
      2] (-(-1 + \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/
         2, -(1/2), \[Rho]^2] + (-1 + (4 + m + 
             n) \[Rho]^2) Hypergeometric2F1[(2 + m)/2, (2 + n)/2, 1/
         2, \[Rho]^2]) (-1 + Cos[m \[Pi]] + I Sin[m \[Pi]]) + 
    Gamma[(1 + m)/2] Gamma[(1 + n)/2] Hypergeometric2F1[(1 + m)/2, (
      1 + n)/2, 1/
      2, \[Rho]^2] (1 + Cos[m \[Pi]] + I Sin[m \[Pi]]) + (-1)^
     n Gamma[(1 + m)/2] Gamma[(1 + n)/2] Hypergeometric2F1[(1 + m)/
      2, (1 + n)/2, 1/
      2, \[Rho]^2] (1 + Cos[m \[Pi]] + I Sin[m \[Pi]]))

$$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\frac{e^{\frac{-x^2-y^2+2 x y \rho }{2 \left(1-\rho ^2\right)}} x^m y^n}{2 \pi \sqrt{1-\rho ^2}}dydx=\frac{2^{\frac{1}{2} (-4+m+n)} \left(1-\rho ^2\right)^{\frac{1}{2} (1+m+n)} \left(\frac{n \Gamma \left(1+\frac{m}{2}\right) \Gamma \left(\frac{n}{2}\right) \left(-\left(-1+\rho ^2\right) \, _2F_1\left(\frac{2+m}{2},\frac{2+n}{2};-\frac{1}{2};\rho ^2\right)+\left(-1+(4+m+n) \rho ^2\right) \, _2F_1\left(\frac{2+m}{2},\frac{2+n}{2};\frac{1}{2};\rho ^2\right)\right) (-1+\cos (m \pi )+i \sin (m \pi ))}{(1+m) (1+n) \rho }+\frac{(-1)^{1+n} n \Gamma \left(1+\frac{m}{2}\right) \Gamma \left(\frac{n}{2}\right) \left(-\left(-1+\rho ^2\right) \, _2F_1\left(\frac{2+m}{2},\frac{2+n}{2};-\frac{1}{2};\rho ^2\right)+\left(-1+(4+m+n) \rho ^2\right) \, _2F_1\left(\frac{2+m}{2},\frac{2+n}{2};\frac{1}{2};\rho ^2\right)\right) (-1+\cos (m \pi )+i \sin (m \pi ))}{(1+m) (1+n) \rho }+\Gamma \left(\frac{1+m}{2}\right) \Gamma \left(\frac{1+n}{2}\right) \, _2F_1\left(\frac{1+m}{2},\frac{1+n}{2};\frac{1}{2};\rho ^2\right) (1+\cos (m \pi )+i \sin (m \pi ))+(-1)^n \Gamma \left(\frac{1+m}{2}\right) \Gamma \left(\frac{1+n}{2}\right) \, _2F_1\left(\frac{1+m}{2},\frac{1+n}{2};\frac{1}{2};\rho ^2\right) (1+\cos (m \pi )+i \sin (m \pi ))\right)}{\pi }$$

Many thanks Mariusz & Jim

(thanks also for the "assumptions" lead. That looks very useful to me!)

There are indeed some explicit results hidden here.

Unfortunately the incomplete version is (with arbitrary lower limits) is vital for what I want to do. What keeps nagging at me (and is the origin of the hunch) is the fact that if you write out a standard MGF type approach, and try and evaluate the "incomplete" expectation value of, say, $$M(u,v)=E[\exp(u x + v y)]=\int_{a}^{\infty}\int_{b}^{\infty} \exp(u x + v y)g(x,y,\rho)dxdy$$ you get a form that looks like with a simple change of variable you can express (I believe) as a simple version of the classical form of the standard, tabulated CDF $$ L(a^{\star},b^{\star},\rho)=f(u,v)\int_{a^{\star}(u,v)}^{\infty} \int_{b^{\star}(u,v)}^{\infty} g(x,y,\rho)dxdy$$. Then you can explicitly evaluate $$(\partial^{2} M/\partial u \partial v)_{ u=0,v=0}$$ etc. to get the desired forms. The algebra looks kinda lengthy so I was hoping to get Mathematica to do the heavy lifting but alas it doesn't want to express results in terms of the CDF as a special function.I was thinking there may be way to define the CDF as my own function then get Mathematica to write stuff in terms of this but a lack of experience with the product left me floundering.

Anyhow, thanks for the pointers!

Tom