Change variables from $x$ to $y = e^{-{1 \over x^2}}$, so that $x = (-\ln y)^{-{1 \over 2}}$ and $dx = {1 \over 2y (-\ln y)^{3 \over 2}}dy$. So the integral becomes
$${1 \over 2} \int_0^{e^{-{1 \over \delta^2}}} e^{i\lambda y} {1 \over y (-\ln y)^{3 \over 2}}dy$$
You can apply the usual stationary phase method here, dividing the integral into $0$ to $f(\lambda)$ and $f(\lambda)$ to $e^{-{1 \over \delta^2}}$ portions for an appropriately chosen $f(\lambda)$. The idea is that $f(\lambda)$ is chosen as small as possible so that integrating by parts in the second integral gives you some improvement if you integrate the $e^{i\lambda y}$ factor and differentiate the ${1 \over y (-\ln y)^{3 \over 2}}$ factor. Since the differentiation gives you a factor of magnitude $C{1 \over |y \ln y|}$ and the integration gives you a factor of magnitude ${1 \over \lambda}$, a natural
choice of $f(\lambda)$ is the $y$ satisfying $|y \ln y| = \lambda^{-1}$. This
can be described in terms of the Lambert $W$ function if desired.

Breaking up the integral into two parts according to this formula, you can bound the first integral by taking absolute values of the integrand and integrating, and bound the second integral by doing the integration by parts, and then taking absolute values of the integrand and integrating. The result of the two integrals should be the same (I think), namely a constant times
$$\int_0^{f(\lambda)} {1 \over y (-\ln y)^{3 \over 2}}\,dy$$
In other words, you get a bound of $C(-\ln f(\lambda))^{-{1 \over 2}}$ for the overall integral. Since $f(\lambda)$ is between $\lambda^{-2}$ and $\lambda^{-1}$ for example, this is the same as a bound of $C(\ln \lambda)^{-{1 \over 2}}$ as in the earlier answer.

I don't know if this bound is optimal since the standard ways of showing such things that I know of don't apply. But this is a nontrivial result given by stationary phase.