Singular asymptotics of Gaussian integrals with periodic perturbations At the bottom of page 5 of this paper by Giedrius Alkauskas it is claimed that, for a $1$-periodic continuous function $f$,
$$
\int_{-\infty}^{\infty} f(x) e^{-Ax^2}\,dx = \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + O(1)
\tag{1}
$$
as $A \to 0^+$.

How can I prove $(1)$?

I'm having a hard time doing it rigorously since I'm unfamiliar with Fourier series.
If I ignore convergence and just work formally then I can get something that resembles statement $(1)$.  Indeed, since $f$ is $1$-periodic we write
$$
f(x) = \int_0^1 f(t)\,dt + \sum_{n=1}^{\infty} \Bigl[a_n \cos(2\pi nx) + b_n \sin(2\pi nx)\Bigr].
$$
Multiply this by $e^{-Ax^2}$ and integrate term by term, remember that $\sin(2\pi nx)$ is odd, and get
$$
\begin{align}
\int_{-\infty}^{\infty} f(x) e^{-Ax^2}\,dx &= \int_{-\infty}^{\infty} e^{-Ax^2}\,dx \int_0^1 f(x)\,dx + \sum_{n=1}^{\infty} a_n \int_{-\infty}^{\infty} cos(2\pi nx) e^{-Ax^2}\,dx \\
&= \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + \sqrt{\frac{\pi}{A}} \sum_{n=1}^{\infty} a_n e^{-\pi^2 n^2/A} \\
&= \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + O\left(A^{-1/2} e^{-\pi^2/A}\right)
\end{align}
$$
as $A \to 0^+$.
I suppose the discrepancy in the error stems from the fact that $f$ need not be smooth (I think in the paper it's actually nowhere differentiable).  Obviously there are some issues, namely the convergence of the series and the interchange of summation and integration.  Resources concerning the analytic properties of Fourier series which are relevant would be much appreciated.
Answers which do not use Fourier series are also very welcome.

Your estimate of error term is correct. The following are just some supplementary details to make your argument more rigorous. 
Let $(f_n)_{n\ge 1}$ be the sequence of partial sums of the Fourier series of $f$, i.e.
$$
f_n(x) = \int_0^1 f(t)\,dt + \sum_{k=1}^n \big(a_k \cos(2\pi kx) + b_k \sin(2\pi kx)\big).
$$
Note that $f_n$ converges to $f$ in $L^2([0,1])$, so by Cauchy-Schwarz's inequality, as $n\to\infty$,
$$\int_0^1|f(t)-f_n(t)| dt\le \big(\int_0^1|f(t)-f_n(t)|^2dt\big)^{\frac{1}{2}}\to 0.\tag{1}$$
Also note that, for every $n\ge 1$,
$$|a_n|=2\cdot\big|\int_0^1 f(t)\cos(2\pi nt) dt\big|\le 2\int_0^1|f(t)|dt.\tag{2}$$
As you have shown,
$$\int_{-\infty}^{+\infty}f_n(x)e^{-Ax^2}dx=\sqrt{\frac{\pi}{A}}(\int_0^1f(t)dt+\sum_{k=1}^na_ke^{\frac{-k^2\pi^2}{A}}).\tag{3}$$
From $(2)$ and $(3)$ we know, 
$$\big|\sqrt{\frac{A}{\pi}}\int_{-\infty}^{+\infty}f_n(x)e^{-Ax^2}dx-\int_0^1f(t)dt\big|\le 2\int_0^1|f(t)|dt\cdot\sum_{k=1}^\infty e^{\frac{-k^2\pi^2}{A}}\le M e^{\frac{-\pi^2}{A}},\tag{4}$$
where $M>0$ is independent of $n\ge 1$ and $0<A\le 1$.
Since for every $m\in\mathbb{Z}$, 
$$\int_m^{m+1}|f_n(x)-f(x)|dx=\int_0^1|f_n(x)-f(x)|dx,$$
$$
\int_{-\infty}^{+\infty}|f_n(x)-f(x)|e^{-Ax^2}dx\le2 \int_0^1|f_n(x)-f(x)|dx\cdot\sum_{m=0}^\infty e^{-Am^2}<\infty.\tag{5}
$$
Letting $n\to\infty$ in $(5)$, from $(1)$ we know that
$$\lim_{n\to\infty}\int_{-\infty}^{+\infty}|f_n(x)-f(x)|e^{-Ax^2}dx=0.\tag{6}$$
Combining $(4)$ and $(6)$, it follows that
$$|\int_{-\infty}^{+\infty}f(x)e^{-Ax^2}dx-\sqrt{\frac{\pi}{A}}\int_0^1f(t)dt|\le M\sqrt{\frac{\pi}{A}} e^{\frac{-\pi^2}{A}}.\tag{7}$$
