I was wondering if $\sum^\infty_{j=a}{\frac{x^{j+(j+b)}}{j!(j+b)!}}$ converges to some function just like the series expansion of $\exp(x)$.
As for the series, $a$ and $b$ are arbitrary non-negative integers.
$$\sum^\infty_{j=a}{\frac{x^{j+(j+b)}}{j!(j+b)!}}=\frac{x^{a+(a+b)}}{a!(a+b)!}+\frac{x^{(a+1)+(a+b+1)}}{(a+1)!(a+b+1)!}+\frac{x^{(a+2)+(a+b+2)}}{(a+2)!(a+b+2)!}+\cdots$$
You can write your sum as
$$\begin{align*}\sum_{i=0}^{\infty}\frac{x^{(i+a)+(i+b)}}{(i+a)!(i+b)!}&=x^{a+b}\sum_{i=0}^{\infty}\frac{(x^2)^i}{(i+a)!(i+b)!}\\ &=\frac{x^{a+b}}{a!b!}\sum_{i=0}^{\infty}\frac{\frac{(i+0)!}{0!}}{\frac{(i+a)!}{a!}\frac{(i+b)!}{b!}}\frac{(x^2)^i}{i!}\\ &=\frac{x^{a+b}}{a!b!}\cdot{}_1\!F_2(1;a+1,b+1;x^2) \end{align*}$$ Where ${}_1\!F_2$ is the Generalized Hypergeometric Function with 1 parameter of type 1 (numerator), and 2 parameters of type 2 (denominator). This isn't really anything profound, since the definition of that function is not very much different to your sum to begin with.
After the edit, some of the parameters are a bit different, but all you really need to do is rewrite $b$ as $a+b$ in the above formulae. You'd obtain $$\frac{x^{2a+b}}{a!(a+b)!}\cdot{}_1\!F_2(1;a+1,a+b+1;x^2)$$