Lets start from the definition of the modified Bessel Function. Recall
$I_{0}(z)=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{z^{2n}}{n!n!}.$ So our integral is
$\int_{0}^{\infty}e^{-\frac{1}{2}\left(x^{2}+b^{2}\right)}x\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{b^{2n}x^{2n}}{n!n!}dx.$ Switching the order of sumation and integration we get
$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx.$ Now, let $u=\frac{1}{2}x^{2}$ and $du=xdx$, to see that $\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=2^{n}\int_{0}^{\infty}e^{-u}u^{n}du=2^{n}\Gamma(n+1)=2^{n}n!.$ Hence
$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\left(2^nn!\right)$
$=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{\left(\frac{b^{2}}{2}\right)^{n}}{n!}=e^{-\frac{1}{2}b^{2}}e^{\frac{1}{2}b^{2}}=1$
as desired.
Notice that the exact same argument shows $\int_0^\infty e^{\left(\frac{1}{2}b^2-\frac{1}{2}x^2\right)}J_0(bx)xdx=1$ where $J_0(x)$ is the Bessel Function.
Hope that helps,