To clarify Chris Heunen's answer, let me point out that most notions of measure theory
have analogs in the category of smooth manifolds.
For example, the analog of a measure space (X,M,μ), where X is a set, M is a σ-algebra of measurable subsets of X, and μ is a measure on (X,M),
is a smooth manifold X equipped with a density μ.
Likewise, the analog of a measure-preserving morphism is a volume-preserving smooth map.
The category of smooth manifolds equipped with a density together
with volume preserving maps as morphisms does not have good categorical properties.
The problem comes from the fact that preservation of volume is too strong a condition to allow for good categorical properties.
If one drops the data of a density and the property of volume preservation,
then the resulting category of smooth manifolds has relatively good categorical properties,
such as existence of finite products and, more generally, existence of all finite transversal limits.
The same is true for the category of measure spaces.
However, in this case we cannot simply drop the data of a measure from the definition
and expect to get a category with good properties.
The reason for this is that the data of a measure in fact combines two independent pieces of data.
The first piece of data tells us which sets have measure zero and which ones don't.
The second piece of data tells us the actual values of measure on sets of non-zero measure.
The analog of dropping the data of a density for measure theory is dropping the second piece
of data described above, but not the first one.
This can already be seen for smooth manifolds: If we have a smooth manifold,
we don't need a density on it to say which sets have measure 0.
Thus one is naturally led to the notion of a measurable space that knows which measurable
sets have measure 0.
I describe it in more detail in [1],
therefore here I offer only a brief summary of the main definitions.
A measurable space is a triple (X,M,N),
where X is a set, M is a σ-algebra of measurable subsets of X, and N⊂M is a σ-ideal
of measure 0 sets.
A morphism of measurable spaces f: (X,M,N)→(Y,P,Q) is an equivalence class of maps of sets g: X→Y
such that the preimage of every element of P is an element of M and the preimage of every
element of Q is an element of N. Two maps g and h are equivalent if they differ on a set of measure 0.
Here for the sake of simplicity I assume that both measurable spaces are complete
(any subset of a measure 0 subset is again a measure 0 subset).
Every measurable space is equivalent to its completion [2],
hence we do not lose anything by restricting ourselves to complete measurable spaces.
In general, one has to modify the above definition to account for incompleteness, as explained
in the link above.
Finally, one has to require that measurable spaces are localizable.
One way to express this property is to say that the Boolean algebra M/N of equivalence
classes of measurable sets is complete (i.e., it has arbitrary suprema and infima).
Many basic theorems of measure theory fail without this property.
In fact, as explained in the link above, theorems such as Radon-Nikodym theorem and Riesz
representation theorem are equivalent to the property of localizability.
Remarkably, the category of localizable measurable spaces is equivalent to the opposite
category of the category of commutative von Neumann algebras [7].
This statement can be seen as another justification for the property of localizability.
Henceforth I assume that all measurable spaces are localizable.
Being in possession of a good category of measurable spaces we can prove that it
admits finite products, and, more generally, arbitrary finite limits.
Let me point out that at this point measure theory diverges from smooth manifolds:
The functor that sends a smooth manifold to its underlying measurable space
is colax monoidal but not strong monoidal
with respect to the monoidal structures given by the categorical products.
More precisely, the category of measurable spaces admits two natural product-like monoidal
structures. One is given by the categorical product mentioned above,
and the other one is the spatial product, which is often simply called the product in many textbooks on measure theory. By the universal property of product there is a canonical map from the spatial product
to the categorical product, which is a monomorphism
but not an isomorphism unless one of the spaces is atomic, i.e., a disjoint union of points.
The forgetful functor F from the category of smooth manifolds to the category of measurable
spaces is strong monoidal with respect to the categorical product on the category
of smooth manifolds and the spatial product on the category of measurable spaces.
Therefore it is also colax monoidal with respect to the categorical product on measurable spaces,
but not strong monoidal because F is essentially surjective on objects and the map
from the spatial product to the categorical product is not always an isomorphism.
Here is an instructive example for the last statement.
Let Z=F(R), where R is the real line considered as a smooth manifold.
Consider the spatial product of Z and Z, which is canonically isomorphic to F(R×R).
The spatial product maps monomorphically to the categorical product Y=Z×Z.
However, Y is much bigger than F(R×R).
For example, the diagonal map Z→Y=Z×Z is disjoint from F(R×R) in Y.
The space Y also has a lot of other subspaces
whose existence is guaranteed by the universal property.
Note that set-theoretically F(R×R) also has a diagonal subset.
However, this subset has measure 0 and therefore is invisible in this formalism.
Let me finish by mentioning that locales arguably provide much better formalism for measure theory,
which in particular does not suffer from problems with sets of measure 0,
e.g., we don't need to pass to equivalence classes or even mention the words “almost everywhere”.
The relevant functor sends a measurable space (X,M,N) to the locale M/N (as explained
above M/N is a complete Boolean algebra, hence a locale).
We obtain a faithful functor from the category of measurable spaces to the category of locales.
Let's call its image the category of measurable locales.
Then measurable morphisms of measurable locales correspond bijectively
to equivalence classes of measurable maps.
Note that we don't need to pass to equivalence
classes to define a measurable morphism of measurable locales.
Every measurable space (or locale) can be uniquely decomposed into its atomic and diffuse part.
The atomic part is a disjoint union of points and the diffuse part does not have any isolated points.
An example of a diffuse space is given by F(M) where M is a smooth manifold of non-zero dimension.
(In fact all these spaces are isomorphic as measurable spaces if the number of connected
components of M is countable.)
Here is the punchline: If Z is a diffuse measurable locale, then it does not have any points,
in particular it is non-spatial.
This can serve as an explanation of why we cannot construct a reasonable concrete category
of measurable spaces and why we always have to use equivalence classes if we want to stay
in the point-set measure theory.
References that advocate (but do not evangelize) the viewpoint described in this answer:
Yes indeed, for fractional $s$ it is related to Modified Bessel Functions of 2nd Kind (Macdonald Function). For $r=|\mathbf{x}|,\ a=|x_{n+1}|$ $$P(r,a)=\frac{a^{2s}}{(r^2+a^2)^{n/2+s}}$$ The Fourier Transform of an n-dimensional radial function $f(r)$ is a radial function in the trasformed space $\mathcal{F}(|\xi|)$. It is given by the following Hankel Transform. $$\mathcal{F}(|\xi|)=\frac{1}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty r^{\frac{n}{2}-1}\left[r\,J_{\frac{n}{2}-1}(|\xi|\cdot r)\right]\cdot f(r)\,dr$$ See, for instance, Sec 11 pg. 63-65 in
Sneddon, Ian N., Fourier transforms, New York: McGraw-Hill Book Company, Inc. (1950). ZBL0038.26801. for a proof.
Also, Sneddon, Ian N., A note on some relations between Fourier and Hankel transforms, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 9, 799-806 (1961). ZBL0100.31501. Therefore $$\mathcal{F}_P(|\xi|,a)=\frac{a^{2s}}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}$$ But integral is formula 6.565.4 in 7-th Edition of
Gradshteyn, I. S.; Ryzhik, I. M., Table of integrals, series, and products. Ed. by Alan Jeffrey. CD-ROM version 1. 0 for PC, MAC, and UNIX computers., San Diego, CA: Academic Press. (1996). ZBL0918.65001.
It reads (adapted to our notation), for $s>0$ $$\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}=\left(\frac{|\xi|}{2}\right)^{\frac{n}{2}+s-1}\cdot\frac{K_s(a\,|\xi|)}{a^s\,\Gamma(\frac{n}{2}+s)}$$ Being $K_s(\cdot)$ the Modified Bessel Function of 2nd Kind. Therefore, $$\mathcal{F}_P(|\xi|,a)=\frac{(a\,|\xi|)^sK_s(a\,|\xi|)}{2^{\frac{n}{2}+s-1}\Gamma(\frac{n}{2}+s)}$$
Since $K_\frac{1}{2}(z)=\sqrt{\frac{\pi}{2z}}\cdot e^{-z}$, for $s=\frac{1}{2}$ we get $$\mathcal{F}_P(|\xi|,a)=\frac{\sqrt{\pi}}{2^\frac{n}{2}\Gamma\left(\frac{n+1}{2}\right)}\cdot e^{-a\,|\xi|}$$ Note: Dimension constant and $2\pi$ in exponent are matter of Fourier Transform's normalization and convention used respectively. By 'Rosetta Table' Formulae 504 at the end of this Wiki page this can be fitted accordingly.
Using question convention -FT unitary, ordinary frequency, $\mathcal{F}^*$- and notation we get,
$$P(|x|,x_{n+1})=\frac{2^{s-\frac{1}{2}}\Gamma\left(\frac{n}{2}+s\right)}{\pi^{\frac{n+1}{2}}}\cdot\frac{x_{n+1}^{2s}}{(|x|^2+|x_{n+1}|^2)^{n/2+s}}$$ producing this amazingly simple expression $$\mathcal{F}_P^*(|\xi|,x_{n+1})=\sqrt{\frac{2}{\pi}}\cdot \left(2\pi\,x_{n+1}\,|\xi|\right)^s\cdot K_s\left(2\pi\, x_{n+1}\,|\xi|\right)$$
This gives $\exp(-2\pi |\xi| x_{n+1})$ for $s=1/2$.
Being so simple, this result should be added as new Formulae 505 in Wiki's Table Fourier Transforms of n-dimensional functions.
I leave this for further reading, a nice article on Fourier Transforms of Fractional Poison Kernels and Fractional Laplacian (See Section 2.8) at
Kwaśnicki, Mateusz, Ten equivalent definitions of the fractional Laplace operator, Fract. Calc. Appl. Anal. 20, No. 1, 7-51 (2017). ZBL1375.47038.
Best Answer
No by the usual duality nonsense. Let's say $\sigma=1$ (it doesn't matter). The convolution at $x$ is $e^{-x^2/2}$ times the integral of $g(t)=f(t)e^{-t^2/2}$ against $e^{tx}$. If the estimate $\left|\int_2^3(f*K)e^{x^2/2}\,dx\right|\le C\int_0^1|f*K|e^{x^2/2}\,dx$ were possible, then there would exist an $L^\infty$ function $h$ on $[0,1]$ such that $\int_{0}^1 h(x)e^{tx}\,dx=\int_2^3 e^{tx}\,dx$ for almost all $t\in[0,1]$, which is clearly impossible (both sides are analytic functions in $t$ and the RHS grows faster than the LHS can afford as $t\to+\infty$).