For this answer, I am going to use Fourier Transform conventions and special function definitions found in Ronald Bracewell's The Fourier Transform and Its Applications.
The 2-D Fourier Transform is defined as:
$$\mathscr{F}\left\{f(x,y)\right\} = F(u,v)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)e^{-i2\pi (ux+vy)}\;dx\;dy$$
The 2-D Fourier Transform can be manipulated into the following Hankel Transform form, when the function has circular symmetry, i.e. when $f(x,y) = f(r)$:
$$\mathscr{F}\left\{f(r)\right\} = F(q) = 2\pi\int_0^\infty f(r)J_0\left(2\pi qr\right)r\;dr$$
where $r$ and $q$ are the radii in the original domain and transform domain respectively, i.e. $r =\sqrt{x^2+y^2}$ and $q = \sqrt{u^2+v^2}$
We'll need the following special function definitions for this answer:
$$\delta(x) \quad \text{(the 1-D Dirac delta function)}$$
$$\Pi\left(x\right) = \begin{cases}
1 & |x| < \frac{1}{2} \\
0 & |x| > \frac{1}{2} \\
\end{cases} $$
$$\mathrm{III}(ax) = \dfrac{1}{|a|}\sum_{n=-\infty}^{\infty} \delta\left(x -\dfrac{n}{a}\right) $$
$$ {}^2 \delta(x,y) = \delta(x)\delta(y) \quad \text{(the 2-D Dirac delta function)}$$
Here are some short descriptions, to give one an intuition of what some applications of these 1-D functions look like when applied in 2-D:
$$\begin{align*} \delta(y) \; &: \;\text{an impulse "wall" running on the x axis} \\
\delta(x)\delta(y) \; &: \;\text{an impulse "spike" at the origin} \\ \mathrm{III}(y) \; &: \; \text{a grate of impulse "walls", each running parallel to the x axis at an integer value of y}\\
\delta(x)\mathrm{III}(y) \; &: \; \text{a row of impulse "spikes", each on the y axis at an integer value of y} \\
\Pi(r) \; &: \; \text{a cylinder of height 1 and radius 1/2 at the origin}\\
\end{align*}$$
I will also mention here, that purely periodic functions can be represented by the convolution of an aperiodic function with a periodic array of impulses.
I am going to define a function $g \; : \mathbb{R}^2 \rightarrow \mathbb{R} $ to represent the function $f$ described in the question, using $\epsilon$ to replace the value $\delta$ in the original question to avoid confusion:
$$\begin{align*} g(x,y) &= 1 - \left[\Pi\left(\dfrac{r}{2\epsilon}\right)*\sum_{n=-\infty}^{\infty}\delta(x)\delta\left(y -{2\pi}n\right)\right]\\
\\
&=1 - \left[\Pi\left(\dfrac{r}{2\epsilon}\right)*\delta(x)\dfrac{1}{2\pi}\mathrm{III}\left(\dfrac{y}{2\pi}\right)\right] \\
\end{align*}$$
where $*$ indicates 2-D convolution.
Taking Fourier Transforms:
$$\begin{align*}\mathscr{F}\left\{g(x,y)\right\} = G(u,v) &= \mathscr{F}\left\{1 - \left[\Pi\left(\dfrac{r}{2\epsilon}\right)*\delta(x)\dfrac{1}{2\pi}\mathrm{III}\left(\dfrac{y}{2\pi}\right)\right]\right\}\\
\\
&= \delta(u)\delta(v) - \mathscr{F}\left\{\Pi\left(\dfrac{r}{2\epsilon}\right)\right\}\mathscr{F}\left\{\delta(x)\dfrac{1}{2\pi}\mathrm{III}\left(\dfrac{y}{2\pi}\right)\right\}\\
\\
&= \delta(u)\delta(v) - \epsilon\dfrac{J_1\left(2\pi\epsilon q\right)}{q}\mathrm{III}\left(2\pi v\right)\\
\\
&= \delta(u)\delta(v) - \epsilon^2\dfrac{J_1\left(2\pi\epsilon q\right)}{2\pi\epsilon q}\sum_{n=-\infty}^{\infty}\delta\left(v -\dfrac{n}{2\pi}\right) \\
\end{align*}$$
Which is an impulse spike of weight 1 at the origin, minus impulse grating slices running parallel to the $u$ axis, of a circular $\mathrm{jinc()}$ function at the origin.
So, for fixed $u$ the Fourier Transform has a line spectrum in $v$, where the weight of the impulses tells one the Fourier coefficients.
I'm not quite sure how to interpret the continuous impulse slices in $u$ for fixed $v$ without some context on the original question.
To plot a visualization, I would set the non-zero value of any $\delta()$ function to $1$.
