The Weierstrass transform$W$ of $f:\mathbb{R}\to\mathbb{R}$ can be defined as the convolution of $f$ with a Gaussian:$$W[f](x) = \frac1{\sqrt{4\pi} } \int_{\mathbb{R}} dy\, f(x-y) e^{-y^2/4}.$$It's not hard to show that the operator $W$ can also be written formally as $W=e^{D^2}$ with $D$ the differentiation operator.Furthermore, the Weierstrass transform can also be applied to distributions and often produces regular functions, for example sending the Dirac delta into a Gaussian.
Another example reported in the Wikipedia page is $f(x)=e^{a x^2}$ with $\mathbb{R}\ni a< 1/4$, which gives via the convolution definition$$W[f](x) = \frac{1}{\sqrt{1-4a}} e^{\frac{ax^2}{1-4a}}.$$The same result can be obtained from $e^{D^2}$ using
$$D^{2n} e^{ax^2} = e^{ax^2} (-a)^{n} H_{2n}(\sqrt{-a} x),%D^{2n} e^{-bx^2} = e^{-bx^2} b^{n} H_{2n}(\sqrt b x) $$with $H_n(x)\equiv (-1)^n e^{x^2}D^n e^{-x^2}$ the Hermite polynomials (this connection is probably not accidental given that Hermite polynomials can themselves be written as $H_n(x)=2^n e^{-D^2/4}x^n$).
Computing the corresponding series should give back $W[f](x)$, using the expansion of a Gaussian in terms of Hermite polynomials:$$\sum_{n=0}^\infty \frac{a^n}{n!} H_{2n}(x) = \frac{1}{\sqrt{1+4a}} e^{\frac{4a}{1+4a} x^2}.$$Numerically it's also easy to see that the expressions coincide... as long as $a\in[-1/4,1/4]$. In other words, we have$$\sum_{n=0}^\infty \frac{D^{2n}}{n!} e^{ax^2} = W[e^{a(\bullet)^2}](x)=\frac{1}{\sqrt{1-4a} } e^{\frac{ax^2}{1-4a}},\quad\forall a\in[-1/4,1/4].$$This is already a good example of what's stated in the page: $e^{D^2}$ doesn't always work as an expression, likely depending on the asymptotic behaviour of the derivatives, here the Hermite polynomials.
So the above shows in a simple case how the convolution and "operatorial" definitions of Weierstrass transform sometimes coincide and sometimes do not.Is there any general result or observation to be made regarding when the expression $e^{D^2}$ actually works?