Prove lower bound of complicated function to show divergence at 0

I'm not sure what kind of tools are available to you, but your original claim follows pretty easily from the monotone convergence theorem, which implies that $$\lim_{\theta \to 0} g(\theta) = \frac{1}{2\pi}\int_0^\infty \frac{e^{-\frac{1}{t}}}{t}\,dt = \infty$$ If you specifically need this lower bound, perhaps another commenter may know better than I do. I suspect you could prove a different lower bound directly using some Taylor expansions and a bunch of calculation (perhaps something like this is worth trying yourself), but I'd rather avoid the details unless it's really necessary.

An explicit way (assuming $\theta^2/4 < 1$): \begin{align} g(\theta) &\geq K \int_1^{\infty}\frac1c \exp \left(-\frac{\theta^2}{4}c \right) \, dc \\ &= K\int_{\theta^2/4}^{\infty}\frac1t \exp \left(-t \right) \, dt \\ & \geq K\int_{\theta^2/4}^{1}\frac1t \exp \left(-t \right) \, dt \\ & \geq \frac{K}{e}\int_{\theta^2/4}^{1}\frac1t \, dt \to \infty. \end{align}

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$$g(\theta) = \frac1{2 \pi} \int_0^\infty \frac{e^{-\frac{\theta^2}4c-\frac1c}}c\, dc\overset{c=\frac{2t}\theta}{=}\frac1{2\pi}\int_0^\infty \frac{e^{-\frac\theta2(t+\frac1t)}}tdt\tag{1}$$ $$\overset{t=e^{z}}{=}\frac1{2\pi}\int_{-\infty}^\infty e^{-\theta\cosh z}dz=\frac1\pi K_0(z)$$ where $\,K_0(\theta)=\int_{-\infty}^\infty e^{-\theta\cosh z}dz$

At $\theta\ll1$ $$K_0(z)=-\ln\theta+\ln2-\gamma+O(\theta^2)$$ $K_0(z)\,\,\text{at}\,\,z\ll1$

(also, here )

Hence, $$g(\theta)=-\frac{\ln\theta}\pi+\frac{\ln2-\gamma}\pi+O(\theta^2), \,\,\theta\ll1$$ The last expression can be also obtained from the integral (1) directly

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If you want good approximation $$K_0(\theta )=\sum_{n=0}^\infty a_n\,\theta^{2n}$$ using $L=-\Big(\gamma+\log \left(\frac{\theta }{2}\right)\Big)$, the first coefficients are $$\left\{L,\frac{L+1}{4},\frac{2 L+3}{128} ,\frac{6 L+11}{13824},\frac{12 L+25}{1769472},\frac{60L+137}{884736000},\frac{20 L+49}{42467328000},\cdots\right\}$$

Using these terms, the relative error is smaller than $0.1$% if $\theta \leq 3$.