I would like to show that given a function $f:\mathbb{R}\to \mathbb{R}$ such that is continuos, $|f(x)| < M$ and $\exists z \in \mathbb{R} $ such that $$-\infty < \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(z + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon < +\infty$$ then:
$$ \forall x,y \in \mathbb{R} \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(x + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon = \lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(y + \varepsilon) \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon $$
I started from the following:
if $$\lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} f(x+\varepsilon) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon = c$$
then we want to show that
$$\lim_{\sigma \to +\infty} \int_{-\infty}^{+\infty} (f(y+\varepsilon)-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon$$
Let $\delta$ be arbitrary then there is $\theta = \theta_{\delta} > 0$ such that
$$\int_{-\theta}^{+\theta} (f(y+\varepsilon)-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon < \int_{-\theta}^{+\theta} (M-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon$$
(\varepsilon is bounded by -\theta and +\theta) and
$$\lim_{\sigma \to +\infty} \int_{-\theta}^{+\theta} (M-c) \frac{1}{2\pi\sigma^2} e^{-\frac{\varepsilon^2}{2\sigma^2}} \mathbb{d}\varepsilon = 0$$
Now It is more difficult to show that the integral between $+\theta$ and $+\infty$ (or the integral between $-\infty$ and $-\theta$ goes to 0. I think in someway the fact that with $x$ it was bounded.
In the present context continuity of $f$ is not important. Rather what happens at $\pm \infty$. Setting $$\rho_\sigma(\varepsilon) = \frac{1}{\sqrt{2\pi} \sigma} \exp\left( - \frac{\varepsilon^2}{\sigma^2}\right) $$
We consider your difference for fixed $x\neq 0$ and with $y=0$: $$ \int_{\Bbb R} \left(f(\varepsilon+x) -f(\varepsilon)\right) \rho_\sigma(\varepsilon) \; d\varepsilon = \int_{\Bbb R} f(\varepsilon) \left( \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon ) \right)\; d\varepsilon$$ We want to show it tends to zero which implies the claim. We have $$ \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon) = \rho_\sigma(\varepsilon) \left[ \exp\left(\frac{2\varepsilon x -x^2}{2\sigma^2}\right)-1 \right] $$ Let $1>\delta>0$ and consider $\epsilon<\sigma^{3/2}$. Then the argument of the exponential is bounded by $$ \left|\frac{2\varepsilon x -x^2}{2\sigma^2}\right|\leq \frac{|x|}{\sqrt{\sigma}} + \frac{x^2}{2\sigma^2}=: a_\sigma$$ which goes to zero as $\sigma\rightarrow \infty$. This implies that $$\int_{-\sigma^{3/2}}^{\sigma^{3/2}}| f(\varepsilon) \left( \rho_\sigma(\varepsilon-x ) - \rho_\sigma(\varepsilon ) \right)|\; d\varepsilon \leq M (\exp(a_\sigma)-1) \rightarrow 0$$ as $\sigma\rightarrow \infty$. For the integral from e.g. $\sigma^{3/2}$ to infinity one shows in a similar way (leaving out the details..., $\rho_\sigma(\varepsilon)$ will be 'very small') that it goes to zero as $\sigma\rightarrow \infty$ whenever $f$ is bounded.