I´m wondering if there exist a vector space A and inner products: $\langle\cdot{,}\cdot\rangle_1$ and $\langle\cdot{,}\cdot\rangle_2$, such that:
An explicit example would be appriciated, Thanks!
Yes. Answering the question as posed in the title instead of the equivalent formulation in the body of the post:
Say $\ell^2(S)=L^2(\mu)$, where $\mu$ is counting measure on $S$. Then $\ell^2(\Bbb N)$ and $\ell^2(\Bbb R)$ give an example. They are certainly not isomorphic, since one has a countable orthonormal basis and the other does not. But they both have algebraic dimension $c$.
First, $\ell^2(\Bbb N)$ is isomorphic to $L^2(\Bbb R)$ (with Lebesgue measure). The set $$\{\chi_{[0,t]}:t>0\}$$is a linearly independent subset of $L^2(\Bbb R)$ of cardinality $c$, so the algebraic dimension of both spaces is at least $c$. But it's not hard to show that $\ell^2(\Bbb R)$ itself has cardinality $c$; hence both spaces have algebraic dimension $c$.
Hint for showing the cardinality of $\ell^2(\Bbb R)$ is no larger than $c$: The cardinality is no larger than the cardinality of $C\times\Bbb R^{\Bbb N}$, where $C$ is the set of countable subsets of $\Bbb R$.