Why does $\gamma=\lim_{s\to1^+}\sum_{n=1}^{\infty}\left(\frac{1}{n^s}-\frac{1}{s^n}\right)=\lim_{s\to0}\frac{\zeta(1+s)+\zeta(1-s)}{2}$?

Dave Renfro's paper ’Euler's Constant $\gamma$' and David Speyer's link should be helpful for an elementary derivation of the first part i.e. get the limit : $$\tag{0}\gamma=\lim_{s\rightarrow 1^+}\left[\zeta(s)-\frac{1}{s-1}\right]$$ As indicated by Gerry your problem was to go from a well defined limit (the limit of the difference $\,\zeta(s)-\frac{1}{s-1}\,$ as $\,s\rightarrow1^+$) to the difference of the limits when these limits are both infinite ! $$-$$ Concerning your second limit : $$\tag{1}\gamma=\lim_{s\to0}\frac{\zeta(1+s)+\zeta(1-s)}{2}$$ this will require a better definition of $\zeta\,$ than $\,\displaystyle\zeta(s):=\sum_{n=1}^{\infty} \frac{1}{n^s}$ since this definition is valid only for $\Re(s)>1$.

To go further you may use the Dirichlet eta function with the idea of converting a sum of positive terms to an alternate sum so that $\zeta$ may then be written as : $$\tag{2}\zeta(s)=-\frac 1{1-2^{1-s}}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}$$ which is convergent for any complex $s$ such that $\,\Re(s)>0,\ s\not =1\,$ or better use the analytic extension of $\zeta$ in the whole complex plane except $s=1$ where $\zeta\,$ has a simple pole (as you found).

To see how to obtain the alternate series $(2)$ (and convergence proof) as well as get some intuitive ideas about analytic continuation of $\zeta\,$ you may see this answer.

Let's note that once the Laurent series of $\zeta$ at $s=1$ obtained with the simple pole at $1$ : $$\tag{3}\zeta(s)=\frac 1{s-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}\gamma_n\;(s-1)^n$$ with $\gamma_n$ the Stieltjes constants and $\gamma_0=\gamma$ your Euler constant then the limit of $\zeta(s)-\frac 1{s-1}$ at $s\to 1$ is rather straightforward.

Using the alternate series or the analytic extension you'll get that the limit was in fact given by (note that $\,s\rightarrow1^+$ was replaced by $\,s\rightarrow1$) : $$\gamma=\lim_{s\rightarrow 1}\left[\zeta(s)-\frac{1}{s-1}\right]=\lim_{z\rightarrow 0}\left[\zeta(1+z)-\frac{1}z\right]$$

The idea is simply to rewrite the $z$ at the right as $+s$ and as $-s$ and to return the mean value to get : $$\gamma=\lim_{s\rightarrow 0}\frac 12\left[\left(\zeta(1+s)-\frac{1}s\right)+\left(\zeta(1-s)-\frac{1}{-s}\right)\right]$$ or $$\gamma=\lim_{s\rightarrow 0}\frac {\zeta(1+s)+\zeta(1-s)}2$$

Let's conclude with an elementary proof using the alternate series $(2)$ :

\begin{align} \zeta(1+s)+\zeta(1-s)&=\sum_{n=1}^{\infty}\frac 1{2^{-s}-1} \frac{(-1)^n}{n^{1+s}}+\frac 1{2^{s}-1} \frac{(-1)^n}{n^{1-s}}\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\frac 1{2^{-s}-1} \frac 1{n^{s}}+\frac 1{2^{s}-1} \frac 1{n^{-s}}\right]\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\left(e^{-s\ln(2)}-1\right)^{-1}e^{-s\ln(n)} +\left(e^{s\ln(2)}-1\right)^{-1}e^{s\ln(n)}\right]\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[2\frac{\ln(n)}{\ln(2)}-1+\sum_{m=1}^\infty s^{2m}\,P_{2m}(\ln(m))\right]\\ \end{align} with $P_{2m}$ polynomials depending of $\ln(n)$ and constants only.
But $\lim_{n\to\infty}\frac{\ln(n)^k}n=0$ for any nonnegative integer $k$ so that we get another nice series equal to $\gamma$ : $$ \lim_{s\rightarrow 0}\frac {\zeta(1+s)+\zeta(1-s)}2=\sum_{n=1}^{\infty}\frac{(-1)^n}n\,\left(\frac{\ln(n)}{\ln(2)}-\frac 12\right)=\gamma$$

As @David Speyer has proven the first part (see link in the comment), this answer focuses on the second part.

As you note, the problem is equivalent to proving: $$\lim_{r\rightarrow1^+}\left(\zeta(r)-\frac{1}{r-1}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right)$$ or equivalently: $$\lim_{s\rightarrow 0^+}\left(\zeta(s+1)-\frac{1}{s}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right) \tag{1}$$ By the reflection formula for the Riemann zeta functionwe have: $$\frac{\zeta \left( s+1 \right)}{2} -{\pi }^{s}\cos \left( 1/2\,\pi \,s \right) \Gamma \left( -s \right) \zeta \left( -s \right) {2}^{s}=0 \tag{2}$$ and so subtracting $(2)$ from $(1)$ we obtain: $$\lim_{s\rightarrow 0^+}\left(\frac{\zeta \left( s+1 \right)}{2}-\frac{1}{s}+{\pi }^{s}\cos \left( 1/2\, \pi \,s \right) \Gamma \left( -s \right) \zeta \left( -s \right) {2} ^{s}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right)$$ and so it remains to prove: $$\lim_{s\rightarrow 0^+}\left(\frac{\zeta \left( 1-s \right)}{2} \right)=\lim_{s\rightarrow 0^+}\left(-\frac{1}{s}+{\pi }^{s}\cos \left( 1/2\, \pi \,s \right) \Gamma \left( -s \right) \zeta \left( -s \right) {2} ^{s}\right) \tag{3}$$ Multiplying $(3)$ by $s$ we obtain: $$\lim_{s\rightarrow 0^+}\left(\frac{s\zeta \left( 1-s \right)}{2}\right)=\lim_{s\rightarrow 0^+}\left(-1+s{\pi }^{s}\cos \left( 1/2\, \pi \,s \right) \Gamma \left( -s \right) \zeta \left( -s \right) {2} ^{s}\right) \tag{4}$$ and $(4)$ holds iff $(3)$ holds. Using the reflection formula again, $(4)$ becomes: $$\lim_{s\rightarrow 0^+}\left(s\cos \left( 1/2\,\pi \,s \right) \Gamma \left( s \right) \zeta \left( s \right) {\pi }^{-s}{2}^{-s} \right)=\lim_{s\rightarrow 0^+}\left(-1+s{\pi }^{s}\cos \left( 1/2\, \pi \,s \right) \Gamma \left( -s \right) \zeta \left( -s \right) {2} ^{s}\right) \tag{5}$$ and we then note that: $$s\Gamma(s)=\Gamma(s+1),\,\,\,-s\Gamma( -s)=\Gamma(-s+1)$$ both of which $\rightarrow 1$ as $s\rightarrow 0$ (as do the other trivial parts of $(5)$), and thus we are left to prove: $$\lim_{s\rightarrow 0^+}\zeta(s)=\lim_{s\rightarrow 0^+}\left(-1-\zeta(-s)\right) \tag{6}$$

The well known limit: $$\lim_{s\rightarrow 0^+}\zeta(s)=-\frac{1}{2}$$ is proven here and the limit is the same from both directions as the Riemann Zeta function is meromorphic; the only pole being at $\zeta(1)$, and thus both sides of $(6)$ are equal to $-1/2$ which proves $(5),(3)$ and hence $(1)$.