Are all series with analytic continuation to a point resulting in Grandi's series equal to 1/2 at that point?
I will first present a proof that two functions $f$ and $g$ whose power series at $l$ and $L$ respcetively are equal (as formal sums) will admit $f(l) = g(L)$.
I will then use this proof to show that the same is true not only for power series, but for taylor expansions in general. If I'm not mistaken, this is all that an analytic continuation really is-- an infinitely long taylor expansion.
I will end with a conclusion addressing your question directly.
I assume throughout this proof that $f$ and $g$ are analytic functions.
Proof that two functions whose power series at points $l$ and $L$ are identical will admit the same values when evaluated at those points:
Suppose we have two functions defined in terms of their power series: $$f(z) = a_0 + a_1 z + a_2 z^2 + \dots$$ $$g(z) = b_0 + b_1 z + b_2 z^2 + \dots$$ We are not concerned with the convergence of these series; think of them as formal sums.
If the terms of the two power series at $l$ and $L$ respectively coincide ($a_n l^2 = b_n L^2 \; \forall n \in \mathbb N_0$) then we have that: $$\frac{a_n}{b_n} = \frac {L^n}{l^n}$$ Now consider that the power series may be equivalently expressed as a series of derivatives of the functions: $$a_n = \frac 1 {n!} \frac {d^n}{dz^n} f(z)$$ $$b_n = \frac 1 {n!} \frac {d^n}{dz^n} g(z)$$ Now let me define a new function $h(z) = f\left(\frac l L z \right)$. Let's look at the power series of this function with coefficients $c_n$: $$c_n = \frac 1 {n!} \frac {d^n}{dz^n} h(z) = \frac 1 {n!} \frac {d^n}{dz^n} f\left(\frac l L z\right) = \frac 1 {n!} \frac {l^n}{L^n} \frac {d^n}{dx^n}f(x)$$ Note something: $$\frac {a_n}{c_n} = \frac {L^n}{l^n}$$ This means that $h(z) = f \left( \frac l L z \right)$ has the same power series as $g(z)$! Thus $g(z) = f \left( \frac l L z \right)$ which in turn implies that $g(L) = f(l)$.
Thus any two functions $f$ and $g$ with equal power series when evaluated at points $l$ and $L$ respectively will have the same analytically continued values $f(l)$ and $g(L)$.
Proof that two functions $f$ and $g$ whose taylor series at points $l$ and $L$ respectively are identical will admit $f(l) = g(L)$:
Suppose now that we expand $f$ around $x$ and $g$ around $y$. I will denote the formal series of $f(z)$ expanded about $x$ as $f_x\{z\}$. $$f_x\{z\} = f(x) + a_1 (z - x) + a_2 (z-x)^2 + \dots$$ $$g_y\{z\} = g(y) + b_1 (z-y) + b_2 (z-y)^2 + \dots$$ Now let $F(z) = f(z + x)$ with $F_0\{z\} = f(x) + c_1z + c_2 z^2 + \dots$
Now $c_n$ is by defintion $\frac {d^n} {dz^n} F(z) |_{z=0}= \frac {d^n} {dz^n} f(z + x) |_{z=0} = \frac {d^n} {dz^n} f(z) |_{z=x} = a_n$. So we have that: $$\forall n, \quad c_n = a_n \quad \quad \Leftrightarrow \quad \quad F_0\{z-x\} = f_x\{z\}$$
We can define $G(z) = g(z +y)$ and come to the same conclusion: $$G_0\{z-y\} = g_y\{z\}$$
Now suppose that $f_x\{l\} = g_y\{L\}$ for some $l,L \in \mathbb C$. Then: $$F_0\{l-x\} = f_x\{l\} = g_y\{L\} = G_0\{L-y\}$$ $$F_0\{l-x\} = G_0\{L-y\}$$
Now we have that the power series of the two functions $F$ and $G$ coincide at $z=l-x$ and $z=L-y$ respectively. I can now make use of the working outlined above and conclude: $$G(z) = F \left( \frac {l-x}{L-y} z \right)$$ $$\Updownarrow$$ $$g(z + y) = f\left(\frac{l-x}{L-y} z + x\right)$$ $$\Updownarrow$$ $$g(z) = f\left(\frac{l-x}{L-y} (z - y) + x \right)$$ From here we can see our answer unfold: $$g(L) = f(l)$$
In conclusion:
So to answer your question directly, I believe that any taylor series expression for a function $f$ at a point on the complex plane which has terms equal to the Grandi series will admit an analytic extension that assigns the value $\frac 1 2$ to that point.
I had been wondering the very same question as you: do we renormalize sums with particular values (i.e. Grandi series $ = \frac 1 2$) because every time that sum occurs in an analytic extension, it is the same? It appears so.
Conveniently, this allows us to assign a value to some non-converging formal sums $ \sum_{n=1}^\infty a_n$ by crafting a function with the power series: $$f(x) = \sum_{n=1}^\infty a_n x^n$$ Analytically extending this function to find the value of $f(1)$ will give us the renormalized value of the sum.
A simple counter-example: $$f_1(s)=1,f_{2n}(s)=-s^{2n},f_{2n+1}(s)=s^{2n-1}$$ for $n\ge1$ and $s_0=1$.
The series converges uniformly on every compact subset of $|s|<1$, to $$1+\frac{s}{1+s}$$
allowing $F$ to be analytically continued beyond $s=1$, and $$F(1)=\frac{3}{2}$$.
The conditions you have given are too weak to uniquely define a value of the sum, in the sense that it allows you to ‘play around’ with which $z^n$ the $n$th $1$ or $-1$ corresponds to.
i.e. $$\sum_{n=0}^\infty a_n z^n=\sum_{n=0}^\infty a_{\sigma(n)} z^n$$ as $|z|\to R$,($\sigma$ is a permutation of $\mathbb N_{\ge0}$) may not hold, in general.