This is one of those precious cases when Ramanujan himself provided (a sketch of) a proof. The identity was published in his paper ["Some definite integrals"][1]  (*Mess. Math.* 44 (1915), pp. 10-18) together with several related formulae. 

It might be instructive to look first at the simpler identity (i.e. the limiting case when $b\to\infty$; the identity mentioned in the original question can be obtained by a similar approach):
$$\int\limits_{0}^{\infty} \prod_{k=0}^{\infty}\frac{1}{  1 + x^{2}/(a+k)^{2}}dx = \frac{\sqrt{\pi}}{2} \frac{ \Gamma(a+\frac{1}{2})}{\Gamma(a)},\quad a>0.\qquad\qquad\qquad(1)$$
Ramanujan  derives (1) by using a partial fraction decomposition of the product  $\prod_{k=0}^{n}\frac{1}{  1 + x^{2}/(a+k)^{2}}$, integrating term-wise, and passing to the limit $n\to\infty$. He also indicates that alternatively (1) is implied by the factorization 
$$\prod_{k=0}^{\infty}\left[1+\frac{x^2}{(a+k)^2}\right] = \frac{ [\Gamma(a)]^2}{\Gamma(a+ix)\Gamma(a-ix)},$$
which follows readily from  Euler's product formula for the gamma function. Thus (1) is equivalent to the formula 
$$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$

------------------------------------------------------------------

There is a nice paper ["Wallis-Ramanujan-Schur-Feynman"][2] by Amdeberhan et al (*American Mathematical Monthly* 117 (2010), pp. 618-632) that discusses interesting combinatorial aspects of formula (1) and its generalizations. 




 


  [1]: http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper11/page1.htm
  [2]: http://arminstraub.com/files/publications/ws.pdf