General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{2x^8}{8!}+...$
$e^x-\sin(x) = 1+\dfrac{x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{2x^7}{7!}+\dfrac{x^8}{8!}+...$
$e^x-\cos(x) = x+x^2+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{2x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$
Due to the missing term and the irregularity, I am unable to write the general formula for these series. I wish to find the general formula with compact sigma notation. Can you help with this?

Take the series for $e^x+\sin x$ for example. This is obtained by taking the series for $e^x$, doubling the terms $\dfrac{x^n}{n!}$ for $n=4k+1$ and deleting the terms $\dfrac{x^n}{n!}$ for $n=4k+3$. So you could write it as
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$
Alternatively, let
$$i_n = \begin{cases} 1 & \text{if } n = 4k \text{ or } 4k+2 \\ 2 & \text{if } n=4k+1 \\ 0 & \text{if } n=4k+3 \end{cases}$$
so that the sequence $(i_n)_{n \ge 0}$ is given by $1,2,1,0,1,2,1,0,\dots$; then you have
$$e^x + \sin x = \sum_{n=0}^{\infty} \dfrac{i_n x^n}{n!}$$
Now note that $i_n = 1 + \sin \dfrac{n\pi}{2}$, to give you
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \left( 1 + \sin \frac{n\pi}{2} \right) \dfrac{x^n}{n!}}$$
You can do a similar trick for the other sequences you desire.
If you don't like using the sine function inside the sum, then you can notice that $i_n = 1 + \dfrac{i^n-i^{-n}}{2}$, where $i^2=-1$; this gives
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \left( 1 + \dfrac{i^n-i^{-n}}{2} \right) \dfrac{x^n}{n!}}$$
If you don't like using the sine function or complex numbers inside the sum, you can do some fiddling around with floor functions and powers of $-1$ to get an equivalent expression. For example:
$$i_n = \dfrac{1-(-1)^n}{2} \cdot (-1)^{\frac{1}{2}\left(n - \left\lfloor \frac{n}{4} \right\rfloor - 1\right)}$$
so you can write
$$\boxed{e^x + \sin x = \sum_{n=0}^{\infty} \dfrac{1-(-1)^n}{2} \cdot (-1)^{\frac{1}{2}\left(n - \left\lfloor \frac{n}{4} \right\rfloor - 1\right)} \cdot \dfrac{x^n}{n!}}$$
