Is the regularization of an otherwise diverging two-sided sum always equal to zero? As a first example, take the divergent series of all powers of two $1+2+4+8+...=\sum\limits_{k=0}^\infty 2^k$ which can be regularized by using the analytical continuation of the geometric series $\sum\limits_{k=0}^\infty q^k = \frac1{1-q}\Big|_{|p|<1}$ to obtain $1+2+4+8+...=-1$, while on the other hand, the sum $\frac12 + \frac14 + \frac18 + ... = 1$, such that $$\sum_{k=-\infty}^\infty 2^k = 0$$
As a second example, take $... -3-2-1+0+1+2+3+...$, which is clearly zero as well (while the half-sided sum requires (Riemann) zeta regularization to obtain $1+2+3+4+...=-\frac1{12}$).
But is this generally the case or did I just pick some exceptional examples?

As a third example - that I am not sure about - take $...+1+1+1+1+...$:
$$\underbrace{...+1+1+1}_{=\zeta(0)=-\frac12} + \underbrace{1}_{\stackrel{\text{from}}{k=0}} + \underbrace{1+1+1+1+...}_{=\zeta(0)=-\frac12} = 0$$ - I am not sure here since I pretend that $\sum_{k=1}^\infty\frac1{(-k)^s}\Big|_{s=0}$ is also $\zeta(0)$ due to the expression's symmetry.

I have a very nice example introducing what I call "alternating iteration series" ("AIS").     
Some years ago I considered the iteration of the function $\small g(x) = \sqrt{x+1/2}$ and its inverse $\small f(x) = x^2-1/2 $. 
Preliminaries
Let's introduce the following notation for convenience: we assume an initial value $x_0$ . We denote an iterate $\small x_0 \to x_1 = g(x_0)$ and in general $\small x_{h+1} = g(x_h)$ . (I use the letter $h$ here because in general I use also the term "h-eight" for the number of iterations, which initially stems from the study of analysis of "power towers" and their visual represention as vertically stacked symbols).
Similarly $\small x_{h-1} = f(x_h)$.     
There are two fixpoints $\small t_f = (1-\sqrt 3)/2$ and $t_g = (1+ \sqrt 3)/2$ so for all initial values $x_0$ in the range $t_f \lt x_0 \le t_g$ the iterates $x_{-\infty}$ and $x_{+\infty}$ converge to that constant values.      
This allows to Cesaro-sum (indicated by the underset $\small \mathfrak C$) the alternating iteration series
$$  \begin{array}{rlll} 
A_{g,p}(x_0) &\underset{\mathfrak C}= x_0^p - x_1^p + x_2^p - ... + ... \\
A_{f,p}(x_0) &\underset{\mathfrak C}= x_0^p - x_{-1}^p + x_{-2}^p - ... + ... \\
\end{array}$$
Even more, once we have that possibility of assigning meaningful values, we can put them together and define
$$ \begin{array}{rlll} A_p(x_0) &= A_{g,p}(x_0) +A_{f,p}(x_0) -x_0^p \\
 &= \cdots - x_{-3}^p +x_{-2}^p-x_{-1}^p + x_0^p - x_1^p + x_2^p -  x_3^p + \cdots
\end{array}$$
Heuristics
Heuristically I found that 
$$ \begin{array}{rlll} A_4(x_0) &= 0 \\ 
&= \cdots - x_{-3}^4 +x_{-2}^4-x_{-1}^4 + x_0^4 - x_1^4 + x_2^4 -  x_3^4 + \cdots
\end{array}$$
for all initial values $\small x_0$ of the allowed range. Moreover, the two partial series can be regularized to simple polynomials in $x$ (which has just recently been proven here in MSE), which also explain that zero-results.       
A generalization - perhaps that still fits your request - are that compositions   of two two-way-infinite series
$$ \begin{array}{rlll}
   A_p(1) + A_p(1/2) &= 0 & \qquad&\text{ for all } p \\
 A_1(x) + A_2(x) &= 0 &&\text{ for all  $x$ in the allowed range}  \\
\end{array}$$
but for which I've no proof yet... 
