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 !
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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 $\gamma$ :
$$
\lim_{s\rightarrow 0}\frac {\zeta(1+s)+\zeta(1-s)}2=\sum_{n=1}^{\infty}\frac{(-1)^n}n\,\left(\frac{\ln(n)}{\ln(2)}-\frac 12\right)=\gamma$$
