Equivalent inner products on a Hilbert space

The first observation to make is that, given a bounded sesquilinear form $B$ on $\mathcal{H}$, i.e., a form such that $|B(x,y)| \leq \|B\|\,\|x\|\,\|y\|$, there exists a unique bounded linear operator $T$ on $\mathcal{H}$ such that $B(x,y) = \langle x,Ty \rangle$ and $\|T\| = \|B\|$ by the Riesz representation theorem (see Pedersen, Analysis Now, Lemma 3.2.2, p.89 for details).

If $B$ is Hermitian, the identity $$\langle x, Ty \rangle = B(x,y) = \overline{B(y,x)} = \overline{\langle y, Tx \rangle} = \langle Tx, y\rangle = \langle x, T^{\ast}y\rangle$$ shows that $T = T^{\ast}$, so $T$ is self-adjoint.

If $B$ is positive definite then of course $T$ must be injective, but it need not be invertible (an injective self-adjoint operator has dense range, of course).

It is not hard to show that your equivalence relation on scalar products implies that the operator $T$ is bounded and bounded away from zero, and as it is self-adjoint, this means that $T$ is invertible, see Pedersen, Proposition 3.2.6, page 90.

Conversely, if $T$ is bounded, self-adjoint and invertible, then $T$ is bounded away from zero and hence the scalar products $\langle x, Ty \rangle$ and $\langle x,y\rangle$ are equivalent.

It is plain that for a scalar product $B$ the condition $$a \langle x,x \rangle \leq B(x,x) \leq b\langle x,x\rangle \quad \text{for all } x \in \mathcal H$$ implies that $\mathcal{H}$ is complete with respect to $B$, so we can sum up:

If a scalar product $B$ satisfies $a \|x\|^2 \leq B(x,x) \leq b \|x\|^2$ for all $x$ then $B(x,y) = \langle x, Ty \rangle$ for a unique bounded, invertible self-adjoint operator $T$. Conversely, for every bounded, invertible self-adjoint operator we get equivalent scalar products.

In particular, there is only one equivalence class of scalar products. It is clear that you can perform all this with bases as well, and, as Willie said, your reasoning is correct. However, these calculations seem more involved to me than the above considerations.

Added:

A closely related result that I should have mentioned is the Lax–Milgram theorem.