$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}\newcommand\C{\mathbb C}$Suppose indeed that $A=c\mathbb Z$ for some $c>0$. By rescaling, the problem reduces to the following: for given real $a$ and $b$ such that $a<b$, recover the (say continuous) function $f\colon[a,b]\to\C$ given the numbers $$\alpha_k:=\int_a^b dy\, f(y) e^{-(Ck-y)^2/2}$$ for some real $C>0$ and all $k\in\Z$. Note that $$\alpha_k=e^{-(Ck)^2/2}\int_a^b dy\, f(y) e^{-y^2/2}e^{Cky}.$$ So, with the substitution $x=e^{Cy}$, the problem reduces to the following inverse moment problem: for given real $s$ and $t$ such that $s<t$, recover the continuous function $h\colon[s,t]\to\C$ given the power moments $$\mu_k:=\int_s^t dx\, h(x) x^k$$ for all $k\in\N_0$. Further, using the substitution $x=s+z$ and the binomial formula, we may assume that here $s=0$.

The resulting inverse moment problem can be solved by an explicit formula, as explained e.g. in the paragraph containing formula (2.2): $$\sum_{0\le j\le ux}\sum_{k=j}^\infty\frac{(-1)^{k-j}u^k}{(k-j)!j!}\,\mu_k\underset{u\to\infty}\longrightarrow\int_0^x dv\,h(v)$$ for all $x\in[0,t]=[s,t]$, as desired. (This formula is stated in the linked paper only for nonnegative $h$. However, this inversion formula holds for complex-valued $h$ by linearity and in view of the decomposition of a real-valued function into its positive and negative parts.)

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}\newcommand\C{\mathbb C}$Suppose indeed that $A=c\mathbb Z$ for some $c>0$. By rescaling, the problem reduces to the following: for given real $a$ and $b$ such that $a<b$, recover the (say continuous) function $f\colon[a,b]\to\C$ given the numbers $$\alpha_k:=\int_a^b dy\, f(y) e^{-(Ck-y)^2/2}$$ for some real $C>0$ and all $k\in\Z$. Note that $$\alpha_k=e^{-(Ck)^2/2}\int_a^b dy\, f(y) e^{-y^2/2}e^{Cky}.$$ So, with the substitution $x=e^{Cy}$, the problem reduces to the following inverse moment problem: for given real $s$ and $t$ such that $s<t$, recover the continuous function $h\colon[s,t]\to\C$ given the power moments $$\mu_k:=\int_s^t dx\, h(x) x^k$$ for all $k\in\N_0$. 

Further, using the substitution $x=s+z$ and the binomial formula, we may assume that here $s=0$. Indeed, letting $H(z):=h(s+z)$ and $T:=t-s$, we have $$\int_0^T dz\, H(z)(s+z)^k=\mu_k.$$ On the other hand, $$z^k=((s+z)-s)^k=\sum_{q=0}^k\binom kq (-s)^{k-q} (s+z)^q.$$ Multiplying the latter identity by $H(z)$ and then integrating in $z$ from $0$ to $T$, we have $$\int_0^T dz\, H(z)z^k=\nu_k:=\sum_{q=0}^k\binom kq (-s)^{k-q} \mu_q.$$ So, now it suffices to recover the function $H$ given its power moments $\nu_k$.

The resulting inverse moment problem can be solved by an explicit formula, as explained e.g. in the paragraph containing formula (2.2): $$\sum_{0\le j\le ux}\sum_{k=j}^\infty\frac{(-1)^{k-j}u^k}{(k-j)!j!}\,\nu_k\underset{u\to\infty}\longrightarrow\int_0^x dz\,H(z)$$ for all $x\in[0,T]$, as desired. (This formula is stated in the linked paper only for nonnegative $h$. However, this inversion formula holds for complex-valued $h$ by linearity and in view of the decomposition of a real-valued function into its positive and negative parts.)