Let $A/\mathbf{F}_q$ be an Abelian variety of dimension $g$. Suppose one knows $|A(\mathbf{F}_{q^n})|$ for all $1 \leq n \leq g$. Does one know then $\zeta(A,s)$ (equivalently, $|A(\mathbf{F}_{q^n})|$ for all $n$)?
It is true for $g = 1$ by an elementary computation.
This is not an answer.
Here is a possible strategy, that was too long for a comment. I briefly thought it gave a full answer, but there is a lot of stuff missing.
If $\alpha_1,\ldots,\alpha_{2g}$ are the eigenvalues on $H^1(A,\mathbb Q_\ell)$, then $\bigwedge^* H^1 = H^*$ computes all eigenvalues. Moreover, any $\frac{q}{\alpha_i}$ is another one of the $\alpha_i$ (see for example [Suh, Prop. 2.2.1]). Thus, the sum $$\#A(\mathbb F_{q^r}) = \sum_{k=0}^{2g} (-1)^k\operatorname{tr}(\operatorname{Frob}_q^r|H^k(A,\mathbb Q_\ell))$$ is entirely determined by suitable algebraic expressions in $\alpha_1,\ldots,\alpha_g$ (w.l.o.g.).
The question then becomes the following: if $f \in \mathbb Q(x_1,\ldots,x_g,y)$ is a rational function symmetric in the first $g$ terms, does knowing $f(\alpha_1^r,\ldots,\alpha_g^r,q^r)$ for $r = 1,\ldots,g$ determine $\alpha_1,\ldots,\alpha_g$ up to permutation?
For $f = x_1+\ldots+x_g$, the answer is yes by classical theory, and it seems plausible that therefore the answer is yes for general $f$. But our specific function may be bad; for example $f = x_1\cdots x_g$ clearly does not have this property.
[Suh] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology, Compos. Math. 148, No. 1, 295-303 (2012). ZBL1258.14023.
SORRY, THIS IS AN ANSWER TO THE WRONG QUESTION
(count of points on a curve instead of on the abelian variety itself)
WILL DELETE OR CORRECT BEFORE LONG
Yes, the counts for $q^n$ ($n \leq g$), together with the value of $q$, are enough.
Let the eigenvalues of Frobenius be $\lambda_i$ ($1 \leq i \leq 2g$), and let $P(t) = \prod_{i=1}^{2g} (1-\lambda t)$ which is a scaling of the characteristic polynomial of Frobenius. Then the power sums $\sigma_n := \sum_{i=1}^{2g} \lambda_i^n$ are the Taylor coefficients of $$ F(t) := -t \frac{P'(t)}{P(t)} = \sum_{i=1}^{2j} \frac{\lambda_i t}{1-\lambda_i t} = \sum_{i=1}^{2j} \left((\lambda_i t) + (\lambda_i t)^2 + (\lambda_i t)^3 + \cdots \right) = \sum_{n=1}^\infty \sigma_n t^n. $$ We are given $\left| A({\bf F}_{q^n}) \right|$, and thus also $\sigma_n = q^n + 1 - \left| A({\bf F}_{q^n}) \right|$, for $n \leq g$.Thus we know $F(t)$ up to $O(t^{g+1})$. Thus we know $-F(t)/t$ up to $O(t^g)$. But that's the logarithmic derivative of $P(t)$, so we know the power series for $\log P(t)$ up to $O(t^{g+1})$ (the constant term vanishes because $P(0)=1$). Thus $$ P(t) = \exp \left(-\!\int_{\tau=0}^t F(\tau) \frac{d\tau}{\tau}\right) $$ gives $P(t)$ up to $O(t^{g+1})$, i.e.\ up to and including the $t^g$ coefficient. Now the functional equation $P(t) = (qt^2)^g P(1/qt)$ gives us the rest: for each $j=1,\ldots,g$, the $t^{g+j}$ coefficient is $q^j$ times the $t^{g-j}$ coefficient. So we know the full expansion of $P$, and thus the full list of eigenvalues and the zeta function, QED.
This is a rambling, unsatisfactory answer, meant to simply sketch what I have learned.
The characteristic polynomial of Frobenius on an abelian variety $A$ of dimension $g$ is of the form $$f(x) = x^{2g}+a_1 x^{2g-1} + a_2 x^{2g-2} + \cdots + a_{g-1} x^{g+1}+a_g x^g$$ $$+q a_{g-1} x^{g-1} + \cdots q^{g-2} a_2 x^2 + q^{g-1} a_1 x + q^g.$$ It has the properties that
(1) all the $a_i$ are integers and
(2) all the roots of $f$ in $\mathbb{C}$ lie on the circle $|z| = q^{1/2}$.
So $f$ is given by $g$ parameters $a_1$, $a_2$, ...,, $a_g$. The theorem of Honda and Tate comes close to saying that any $a_i$ obeying (1) and (2) give the characteristic polynomial of a $g$-dimensional abelian variety. Actually, what it actually says is that any such $f$ divides the characteristic polynomial of some abelian variety, possibly of larger dimension, but in many cases (for example, whenever $a_g \not \equiv 0 \bmod p$) the dimension is $g$, so lets just look at (1) and (2).
The number of points of $A(\mathbb{F}_{q^r})$ is $N_r := \prod_{\zeta^r=1} f(\zeta)$. This is clearly a polynomial in the $a_i$, and it isn't bad to show that the values for $1 \leq r \leq g$ are algebraically independent. So, given generic values of $N_1$, $N_2$, ..., $N_r$, there are only finitely many values of $a_i$ in $\mathbb{C}^g$ giving rise to them. It is likely that not all of these solutions obey (1) and (2).
We note that it is natural to instead work with the product only over the primitive $r$-th roots of $1$, which I'll term $N'_r$. So $N_r = \prod_{s|r} N'_s$. Since $N'_r$ is a polynomial of degree $\phi(r)$ (the Euler totient function) in the $a_i$, it seems natural to conjecture that the field of formal rational functions $\mathbb{Q}(a_1, \ldots, a_g)$ is degree $\prod_{i=1}^g \phi(i)$ over the subfield $\mathbb{Q}(N'_1, \ldots, N'_g)$, and hence over $\mathbb{Q}(N_1, \ldots, N_g)$. But I didn't find a statement of this recorded, and it wasn't obvious to me how to prove it.
As the OP notes, for $g=1$, the quantity $f(1) = q+1-a_1$ determines $a_1$. Also, when $g=2$, the quantities $N'_1 = f(1) = q^2+1 + (q+1) a_1 + a_2$ and $N'_2 =N_2/N_1 = f(-1) = q^2+1 - (q+1) a_1 + a_2$ determine $a_1$ and $a_2$. (Since $A$ is an abelian variety, it has an identity in $A(\mathbb{F}_q)$, so $N_1 \geq 1$ and we may divide by it.) So the first interesting case is $g=3$.
Searching for a counter-example If I disregard (1) and (2), it is easy to give examples of nonuniqueness with $g=3$. Let $\omega$ be a primitive cube root of $1$. Find a real polynomial $h$ of the form $$h(x) = x^6+a_1 x^5+a_2 x^4+a_3 x^3 + q a_2 x^2 + q^2 a_1 x + q^3$$ such that $$\mathrm{arg}(h(\omega))+\pi/2 = \mathrm{arg}(\omega (\omega^2-1)(\omega^2-q)).$$ Choose some real scalar $t$ and set $f_{\pm}(x) = h(x) \pm t x (x^2-1)(x^2-q)$. Then $$f_+(1)=f_-(1)=h(1)$$ $$f_+(-1)=f_-(-1)=h(-1)$$ $$f_+(\omega) f_+(\bar{\omega}) = |h(\omega)|^2 + t^2 |\omega (\omega^2-1)(\omega^2-q)|^2 = f_-(\omega) f_-(\bar{\omega}).$$ So $f_+(1)=f_-(1)$, $f_+(-1)=f_-(-1)$ and $f_+(\omega) f_+(\bar{\omega})=f_-(\omega) f_-(\bar{\omega})$ but $f_+ \neq f_-$.
It is easy to find such $f$ satisfying (1) or (2).
But I haven't been able to get both to hold at once! I searched through all degree 6 Weil polynomials for $q \leq 11$ without finding an example, using the formulas here. Indeed, heuristically, I'm not sure I expect a collision. The Weil polynomials are the lattice points $(a_1, a_2, a_3)$ in $\mathrm{diag}(q^{1/2}, q, q^{3/2}) \cdot U$, where $U$ is the set of $(b_1,b_2,b_3) \in \mathbb{R}^3$ such that $x^6+b_1x^5+b_2 x^4+b_3 x^3 + b_2 x^2+b_1 x + 1$ has roots on the unit circle. So the number of Weil polynomials should grow like $q^{1/2} q q^{3/2} = q^3$. Meanwhile, the values of $f(1)$ and $f(-1)$ are of order $q^3$, and $f(\omega) f(\bar{\omega})$ is of order $q^{12}$. Since $(q^3)^2 \ll q^{12}$, the Birthday paradox suggests no collisions. I'd be interested to see if anyone can find one.
Sturmfels and Zworski have an unpublished conjecture, the "Chez Panisse conjecture", which I believe is meant to say that that adding one more $N_r$ to the list will determine $f$. I say "I believe" because the published restatements I can find (Hillar's NSA proposal, Hillar and Levine, Kedlaya) state the conjecture as saying that, if $f$ is reciprocal (meaning $f(x)=x^{2g} f(1/x)$) then $f$ can be recovered from its first $g+1$ cyclic resultants (meaning $\prod_{\zeta^r=1} f(\zeta)$). But $f$ for an abelian variety obeys $f(x)=x^{2g} q^{-g} f(q/x)$, not $f(x)=x^{2g} f(1/x)$ and, while we could make a change of variable to work with a reciprocal $f$, I believe we would then want $\prod_{\zeta^r=1} f(q^{-1/2} \zeta)$. Since I can't find a place where Sturmfels and Zworski themselves wrote this down, I can't say whether they missed this issue, whether everyone reporting on them did, or whether they for some reason wanted to state a conjecture which wasn't quite the right one for applications to number theory.
The above cited work of Kedlaya shows that $\max(2g,18)$ values suffice, where we are allowed to use conditions (1) and (2) to eliminate solutions.
In case anyone would like Mathematica code to generate a complete list of Weil polynomials of degree $6$, here you go:
wPolys[q_] :=
Flatten[Table[
t^6 + a1 t^5 + a2 t^4 + a3 t^3 + q a2 t^2 + q^2 a1 t + q^3,
{a1, Ceiling[-6 Sqrt[q]], Floor[6 Sqrt[q]]},
{a2, Ceiling[4 Sqrt[q] Abs[a1] - 9 q], Floor[a1^2/3 + 3 q]},
{a3,
Ceiling[Max[
-2 q a1 - 2 Sqrt[q] a2 - 2 q^(3/2), -2 a1^3/27 + a1 a2/3 +
q a1 - (2/27) (a1^2 - 3 a2 + 9 q)^(3/2)]],
Floor[Min[
-2 q a1 + 2 Sqrt[q] a2 + 2 q^(3/2), -2 a1^3/27 + a1 a2/3 +
q a1 + (2/27) (a1^2 - 3 a2 + 9 q)^(3/2)]]}]]
The number of such for the primes $2$, $3$, $5$, $7$, $11$ is $215$, $677$, $2953$, $7979$, $30543$.