The answer to your question is yes - by now you've probably already seen this and moved on, but I post this anyway for the benefit of other readers. The hyperbolic functions, such as
$\sinh$ and $\cosh$,
are one of those topics that is almost always skipped right over in undergraduate math, then if it happens to pop up after that everyone acts like you must have seen it a million times. As a result there seem to be a lot of questions like this on the web, often with overly complicated answers. Anyway, here's my attempt at aiding that situation.
The hyperbolic functions are so called because they arise from the hyperbola
$x^2-y^2=1$, analogously to how the ordinary trigonometric functions relate to the circle
$x^2+y^2=1$
(but not quite in the way you might think).
Recall how if
$\theta$ is an angle off the positive $x$-axis in $\mathbb{R}^2$, going counterclockwise, then the point on the unit circle intercepted by $\theta$ will be $(\cos\theta, \sin\theta)$. We want to do something similar, replacing the circle with the hyperbola as mentioned, but for the analogy to work we need to describe the unit circle scenario a little differently.
First of all, get a good picture in your head (or draw one, or look one up) of the hyperbola
$x^2-y^2=1$.
It has two components: one on either side of the $y$-axis, and if we consider the places where an angle $\theta$ as above would intercept a point on it, we would be limited to
$\theta\in(-\frac{\pi}{4},\frac{\pi}{4})\cup(\frac{3\pi}{4},\frac{5\pi}{4})$
plus multiples of
$2\pi$.
But as you can see from the exponential definition of these functions, $\cosh$ and $\sinh$ have domain $\mathbb{R}$.
So, instead of thinking about the angle $\theta$, think about the area $\alpha$ bounded by its initial ray (the $x$-axis), its terminal ray, and the circle. Putting $\theta$ in radians, this area is $\alpha=\theta/2$ (for instance, the area of the whole circle is $\pi$, which is traversed by the full rotation $2\pi$). Looking now at the point intercepted by $\theta$ on the circle in terms of $\alpha$, we have $(\cos 2\alpha,\sin 2\alpha)$.
This is the correct setup for moving to the hyperbolic setting. Suppose $\alpha$ is now the area bounded by the $x$-axis, some other ray $\rho$ coming out of the origin, and the hyperbola $x^2-y^2=1$. Now identify the point on the hyperbola intercepted by $\rho$. The coordinates of this point will be $(\cosh 2\alpha, \sinh 2\alpha)$. Note that the $\alpha$ values you can get in this way are arbitrarily large, even though to bound an area $\rho$ can't pass (or hit) $\frac{\pi}{4}$ off the $x$-axis - using some calculus (and probably an integral table - the integral is ugly) you can see that the area between the diagonal line and the hyperbola, in the first quadrant, is infinite. If you want negative $\alpha$, just move your ray down and call the area negative.
This gives your second identity easily. The first one takes a little more work.
Unless you're forced to use power series, I don't think that power series are the best way to solve this problem. However, if you really want to use power series, note that we can begin by simplifying the integrand using the laws of logarithms to get
$$
\int_0^1\ln\frac{1}{1-x}dx=-\int_0^1\ln(1-x)dx.
$$
Next, observe that this integral is an improper integral since $\ln(1-1)=\ln(0)$ which is undefined. Therefore, we rewrite this as
$$
-\lim_{r\rightarrow 1^-}\int_0^r\ln(1-x)dx.
$$
Note that $\ln(1-x)$ has Taylor series expansion $-\sum_{n=1}^\infty\frac{x^n}{n}$ centered at $0$ with radius of convergence $1$. Therefore, for any $r$ considered above, we can substitute the Taylor expansion to get
$$
\lim_{r\rightarrow 1^-}\int_0^r\sum_{n=1}^\infty\frac{x^n}{n}dx.
$$
Since Taylor (power) series are uniformly convergent within their interval of convergence (this can be made more precise or ignored, depending on your level), we can interchange the limit and the derivative to get that this equals
$$
\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\int_0^r\frac{x^n}{n}dx=\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r.
$$
Once again, using uniform convergence, we can exchange the sum and the limit to get
$$
\sum_{n=1}^\infty\lim_{r\rightarrow 1^-}\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r=\sum\lim_{r\rightarrow 1^-}\frac{r^{n+1}}{n(n+1)}=\sum_{n=1}^\infty\frac{1}{n(n+1)}.
$$
Finally, we can use a partial fraction decomposition to write this as
$$
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right),
$$
which telescopes to $1$.
Now, if I were solving this without any restrictions, I would take the integral
$$
-\int_0^1\ln(1-x)dx
$$
and use the $u$-substitution $u=1-x$ so $du=-dx$ to get
$$
\int_1^0\ln(u)du.
$$
This integral is, once again, improper when $u=0$, so we write
$$
\lim_{r\rightarrow 0^+}\int_1^r\ln(u)du.
$$
An antiderivative of $\ln(u)$ is $u\ln(u)-u$, which can be found through integration by parts, for example. Therefore, this simplifies to
$$
\lim_{r\rightarrow 0^+}(u\ln u-u)|_1^r=\lim_{r\rightarrow 0^+}(r\ln r-r)-(1\ln 1-1).
$$
Now, $\lim_{r\rightarrow 0^+}r\ln r$ is an indeterminate form, but an application of l'Hopital's rule results in a value of $0$, so this entire expression simplifies to $1$, as expected.
Best Answer
You're referring to a multisection of a series. In your case, where you split it up into three functions $E_0(x),E_1(x),E_2(x)$, you get the system: $$ \begin{cases} E_0(x)+E_1(x) + E_2(x) = e^x\\ E_0(0)=1;\qquad E_0'(0)=0\\ E_0'(x) = E_1(x);\qquad E_1'(x)= E_2(x);\qquad E_2'(x)=E_0(x) \end{cases} $$In this case, you get $E_0(x) = \frac{1}{3}\left(e^x+2 e^{-x/2}\cos\left(\frac{\sqrt{3}}{2} x\right)\right)$.