Hilbert-Poincaré Series of Finite-Dimensional Graded Algebras Suppose I have two finite-dimensional $\mathbb{Z}_{\geq 0}$-graded $\mathbb{C}$-algebras $A = \bigoplus_{k \geq 0} A_{k}$ and $B = \bigoplus_{k \geq 0} B_{k}$ with Hilbert-Poincaré series, $P_{A}(t) = \sum_{k \geq 0} \dim A_{k} \ t^{k}$ and $P_{B}(t) = \sum_{k \geq 0} \dim B_{k} \ t^{k}$, respectively.
When is it true that $A$ and $B$ are isomorphic as graded $\mathbb{C}$-algebras if $P_{A} = P_{B}$? Suppose that $P_{A} \neq P_{B}$ but $P_{A}(1) = P_{B}(1)$, what can be said about $A$ and $B$ in this case? Are they isomorphic as $\mathbb{C}$-algebras but not as graded $\mathbb{C}$-algebras?
The algebras that brought me to ask these questions are all of the form $\mathbb{C} \{ z_1, \dots, z_n \} / J$, where $J$ is a finitely generated ideal of partial derivatives of a complex analytic function $f$ with an isolated critical point at the origin.
 A: Your claim that equality of Hilbert-Poincaré series implies equality of algebras is false.        
Take $A=\mathbb C [X,Y]/(X^2)$
and   $B=\mathbb C [X,Y]/(X^2+Y^2)$
with their quotient  gradings.    
We have $P_{A}(t)=P_{B}(t) =1+2t+2t^2+2t^3+2t^4+\ldots=\frac{1+t}{1-t}$ .
But $A$ and $B$ are not isomorphic as rings (and thus certainly not as graded algebras), since $A$ has a non-zero nilpotent element (the class of $X$),  whereas $B$ is reduced i.e. has zero as only nilpotent element. 
Generalization If you consider a homogeneous polynomial of degree $d$ in $n$ variables $f(X_1,\ldots,X_n)\in \mathbb C [X_1,\ldots,X_n]$, then the algebra 
$A(f)=\mathbb C [X_1,\ldots,X_n]/\lt f(X_1,\ldots,X_n)\gt$ has a Hilbert-Poincaré series independent of $f$, namely (combinatorists: please combine!)
$$P(t)=\sum_{k\geq 0}\binom{k+n-1}{n-1}t^k-\sum_{k\geq d}\binom{k-d+n-1}{n-1}t^k$$       
However the rings $A(f)$ strongly depend on $f$. This is evident geometrically because the hypersurfaces $Z(f)=\{ x \in  \mathbb C^n|f(x)=0\}$ vary a lot with $f$: they may be irreducible or not, smooth or not, etc. And so their rings of functions $A(f)$ differ a lot too and are not in general  isomorphic for different $f\;$'s.
Edit I had overlooked that the OP wanted finite-dimensional examples. It is easy to modify the above examples by killing off all monomials of sufficiently high degree. Geometrically, this means looking at the intersection or the hypersurface $Z(f)$ with a sufficiently large infinitesimal neighbourhood of the origin. Let me do this explicitly for the first example and kill all monomials of degree $\geq 3$:
Put  $A'=\mathbb C [X,Y]/(X^2,XY^2,Y^3)=\mathbb C [x,y ]\;$ and  $B'=\mathbb C [U,V]/(U^2+V^2,U^3,V^3)=\mathbb C [u,v]$  .
Then  $P_{A'}(t)=P_{B'}(t) =1+2t+2t^2 \;$.
However the algebras $A'$ and $ B'$are not isomorphic: for example, their spaces of solutions of the equation $\xi^2=0$  have dimensions $3$ and $2$ respectively over $\mathbb C$. Namely,  these spaces are:
$Nil_2(A')=\mathbb C . x\oplus \mathbb C .xy \oplus \mathbb C .y^2$ and $Nil_2(B')=\mathbb C . u^2 \oplus \mathbb C .uv$
A: Asking "does $P_A=P_B$ imply $A \cong B$?" is like asking "does $|G|=|H|$ imply $G \cong H$?": the answer is no, except in certain very special  circumstances.  If $P_A = 1$, or $1+t^{57}$, or $1+t^2+t^7$ etc. the answer is yes, but in general it is no: given a series $P_A$ simply construct an algebra with the same graded pieces as $A$ but with trivial multiplication.  This will not normally be isomorphic to $A$.  
If $P_A(1) = P_B(1)$ you have much less information.  If they are both 2 you can say something, but I doubt you can get much more without restricting the class of algebras
A: The simplest counterexample is, I think, this:
$$A=A_0=\mathbb C\times\mathbb C,\quad B=B_0=\mathbb C[X]/(X^2).$$
