Noticing that $$a_{k,n} = \frac1{\prod_{i=1}^n (k+2i)} = \frac1{2^n(n-1)!}\int_0^1 (1-t)^{n-1}t^{k/2}\,{\rm d}t,$$ we get $$P_n(x) = \frac{e^{-x}}{2^n(n-1)!} \int_0^1 (1-t)^{n-1} e^{x\sqrt{t}}\,{\rm d}t.$$ Maple computes intergral in terms of modified Bessel $I$ and Struve $L$ functions, giving $$P_n(x) = e^{-x}\bigg(\frac1{2^n n!} + \frac{\sqrt{2\pi}}{2}(-x)^{-n+\frac12}(L_{n+\frac12}(-x)-I_{n+\frac12}(-x))\bigg).$$
Alternatively, we can use binomial expansion and substitution $t=z^2$ to get: \begin{split} P_n(x) &= \frac{e^{-x}}{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \int_0^1 z^{2k+1} e^{xz}\,{\rm d}z \\ &=\frac1{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k x^{-(2k+2)} (2k+1)! \bigg(e^{-x} + \sum_{i=0}^{2k+1} (-1)^{i+1} \frac{x^i}{i!}\bigg) \\ &=\frac{e^{-x}}{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k x^{-(2k+2)} (2k+1)! \\ &\quad + \frac1{2^{n-1}(n-1)!} \sum_{j=1}^{2n} (-1)^{j+1} (j-1)! x^{-j} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \binom{2k+1}{j-1}. \end{split}
Using Euler transformation, theThe inner sum can be evaluated as follows $$\sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \binom{2k+1}{j-1} = (-1)^{n-1} 2^{2n-j-1} \frac{j}n \binom{n}{j-n},$$\begin{split} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \binom{2k+1}{j-1} &= \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k [z^{j-1}]\ (1+z)^{2k+1} \\ &= [z^{j-1}]\ (1+z) (1-(1+z)^2)^{n-1} \\ &= (-1)^{n-1} [z^{j-n}]\ (1+z) (2+z)^{n-1} \\ &= (-1)^{n-1} \bigg(\binom{n-1}{j-n} 2^{2n-j-1} + \binom{n-1}{j-n-1} 2^{2n-j} \bigg)\\ &=(-1)^{n-1} 2^{2n-j-1} \frac{j}n \binom{n}{j-n}, \end{split} which allows us to simplify the polynomial part of $P_n(x)$ to $$2^n\sum_{j=n}^{2n} (-1)^{n+j} \frac{j!}{(j-n)!(2j-n)!} (2x)^{-j}.$$