Let $\varphi_\mu$ denote the characteristic function of an infinitely divisible probability measure $\mu$ on $\mathcal B(\mathbb R^d)$ and $\mu^{\ast t}$ denote the convolution power for $t\ge0$. We can show that there is a unique continuous $f:\mathbb R^d\to\mathbb C$ with $f(0)=0$ and $$\varphi_\mu=e^f\tag1$$ and hence $$\varphi_{\mu^{\ast t}}=e^{tf}\tag2.$$
Now let $$(\kappa_tg)(x):=\int g(x+y)\:\mu^{\ast t}({\rm d}y)\;\;\;\text{for }x\in\mathbb R^d$$ for bounded Borel measurable $g:\mathbb R^d\to\mathbb C$ and $t\ge0$.
Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. We can show that if $g\in\mathcal L^1(\lambda^{\otimes d})$ is bounded, then the Fourier transform of $\kappa_tg$ is given by $$\mathcal F(\kappa_tg)=e^{tf}\mathcal Fg\tag3$$ for all $t\ge0$.
Assume $g$ is contained in the Schwarz space $\mathcal S(\mathbb R^d;\mathbb C)$ and let $$p(t,\;\cdot\;):=e^{tf}\mathcal F g\;\;\;\text{for }t\in\mathbb R.$$
Since the Schwarz space is preserved under the Fourier transform and $\mathcal S(\mathbb R^d;\mathbb C)\subseteq\mathcal L^1(\lambda^{\otimes d};\mathbb C)$, we obtain that $$p(t,\;\cdot\;)\in\mathcal L^1(\lambda^{\otimes d};\mathbb C)\tag4\;\;\;\text{for all }t\ge0.$$
Moreover, we easily see that $p(\;\cdot\;,x)$ is differentiable for all $x\in\mathbb R^d$ and $$\frac{\partial p}{\partial t}(t,\;\cdot\;)=fp(t,\;\cdot\;)\;\;\;\text{for all }t\ge0\tag5.$$
My goal is to show that $$\frac{\kappa_tg-g}t\xrightarrow{t\to0+}(\mathcal F^{-1}(f\mathcal Fg))\tag6,$$ where $\mathcal F^{-1}$ denotes the inverse Fourier transform.
We could do so by differentiation under the integral sign. However, in order to apply this theorem, we need that there is a $h\in\mathcal L^1(\lambda^{\otimes d})$ with $$\left|\frac{\partial p}{\partial t}(t,\;\cdot\;)\right|\le h\tag6.$$
If we could show $(6)$, we could conclude that, for fixed $y\in\mathbb R^d$, $$\frac{\partial p}{\partial t}(t,\;\cdot\;)\in\mathcal L^1(\lambda^{\otimes d};\mathbb C)\tag7\;\;\;\text{for all }t\ge0$$ and $$[0,\infty)\times\mathbb C\;,\;\;\;t\mapsto\left(\mathcal F^{-1}p(t,\;\cdot\;)\right)(y)\tag8$$ is differentiable with $$\frac\partial{\partial t}\left(\mathcal F^{-1}p(t,\;\cdot\;)\right)(y)=\left(\mathcal F^{-1}\frac{\partial p}{\partial t}(t,\;\cdot\;)\right)(y)\tag9.$$
In particular, for $t=0$, $$\lim_{h\to0+}\frac{(\mathcal F^{-1}\mathcal F(\kappa_tg))(y)-g(y)}t=\left.\frac\partial{\partial t}\left(\mathcal F^{-1}p(t,\;\cdot\;)\right)(y)\right|_{t=0}=\left(\mathcal F^{-1}\left.\frac{\partial p}{\partial t}(t,\;\cdot\;)\right|_{t=0}\right)(y)=(\mathcal F^{-1}(f\mathcal Fg))(y)\tag{10}$$ for all $y\in\mathbb R^d$.
So, besides verifying $(6)$, we additionally need to show that $\mathcal F^{-1}\mathcal F(\kappa_tg)=\kappa_t$ for all $t\ge0$. In order to this, it would be sufficient to show that $\kappa_tg\in\mathcal S(\mathbb R^d;\mathbb C)$.
Remark: Note that we can show that $\sqrt{|f|}$ is subadditive and hence $$|f(y)|\le c(1+\left\|y\right\|^2)\tag{11}\;\;\;\text{for all }y\in\mathbb R^d$$ for some $c\ge0$.