Let $a_i=(h_{\gamma i})$ be a sequence in $H$.
Assume $A=\sum_i \| a_i \| < \infty $. So $A=\sum_i \sqrt{\sum_\gamma \|h_{\gamma i} \|^2 } \geq \sum_i \|h_{\gamma i} \|^2 $. So for each $\gamma$, $g_\gamma=\sum_i h_{\gamma i}$ converges.
Now let $a=(g_\gamma)$, then
$$
\infty > A^2 =\sum_{i,j} \sqrt{\sum_\gamma \| h_{\gamma i} \|^2}\quad \sqrt{\sum_\gamma \| h_{\gamma j} \|^2}\geq \sum_{i,j}\sum_\gamma \| h_{\gamma i} \| \| h_{\gamma j} \| \geq
$$
$$ \sum_{i,j}\sum_\gamma \langle h_{\gamma i}, h_{\gamma j} \rangle =\sum_\gamma \| h_{\gamma i} \|^2 = \|a\|
$$
is finite, so $a\in H$ and $\sum a_i$ converges to $a$.
Here is a very general, abstract construction. Let $X$ be an incomplete seperable inner product space and let $Y$ be its completion. Since $X$ is incomplete, there is some $y \in Y \setminus X$. If we had $y \perp X$, this would imply (by density) that $y \perp Y$ and hence $y = 0\in X$, a contradiction. Hence, $y\not\perp X$. Thus, by renormalizing, there is $x_0 \in X$ with $\langle x_0, y\rangle =1$.
Now, the function $\varphi : Y \to \Bbb{K}, x \mapsto \langle x, y\rangle$ is a bounded functional on $Y$ and thus restricts to a bounded linear functional on $X$. Let
$$
M := \{x \in X \mid \varphi(x) = 0\} = X \cap \rm{ker}\,\varphi
$$
and note that $M$ is closed in $X$.
Furthermore, $M \subset \rm{ker}\,\varphi$ is dense. Indeed, let $z \in \rm{ker}\,\varphi$. Then there is a sequence $(x_n)_n$ in $X$ with $x_n \to z$. Hence, $\varphi(x_n) \to \varphi(z) = 0$. Now let $x_n ' := x_n - \varphi(x_n) x_0$. Then $x_n ' \in M$ and it is easy to see $x_n \to z$.
Now, choose a countable dense set $(m_n)_n$ in $M$ and orthonormalize it using the Gram Schmidt procedure, producing an orthonormal set $(x_n)_n$ in $M$ with
$$
\overline{\rm{span}(x_n)_n} = \overline{\rm{span}(m_n)_n} = M.
$$
Here, we take the closure in $X$. Note that the above is indeed true (we don't get all of $X$), since $M$ is closed in $X$. Thus, $(x_n)_n$ is not an orthonormal basis of $X$.
But $(x_n)_n$ is complete in your sense: Since if $x \in X$ satisfies $x \perp x_n$ for all $n$, then (by density) $x \perp M$. But as saw above, $M$ is dense in $\rm{ker}\,\varphi$, so that $x \perp \rm{ker}\,\varphi$ (where we consider $x$ as an element of the completion $Y$).
But it is easy to see $\rm{ker}\,\varphi = (\rm{span}(y))^\perp$, so that (recall that $Y$ is complete) we get $x \in ((\rm{span}(y))^\perp)^\perp = \rm{span}(y)$. Since $y \in Y \setminus X$ and $x \in X$, this implies $x=0$.
One can now certainly make this construction concrete by choosing e.g. $X = \Bbb{K}[X]$ and $Y = L^2([0,1])$ or $X = \ell_0$ (the finitely supported sequences) and $Y = \ell^2$, but I leave this to you as an exercise.
Best Answer
There's a notion of completeness of a metric space and a notion of completeness of a basis, and they're not the same thing. Hilbert spaces are defined to be, in particular, complete metric spaces. Completeness of a basis means something different. It means what you said, and another way of stating it is that the span of the given system is dense in the Hilbert space. So, one is an intrinsic property of the space itself and the other is a property of particular subspaces.
For the record, I think a more clear way to state completeness of a system than you did is that if $\langle g,\varphi_j\rangle=0$ for all $j$, then $g=0$.