I think it's not the case. Put $A_n(e_k)=0$ if $k\neq n$ and $A_n(e_n)=ne_n$. Then $A_n$ is a bounded linear operator with norm equal to $n$, and for all $h,k\in\mathbb N$, $A_ne_k=0$ if $n\geq k+1$, so the property $(e_h, A_ne_k)\to (e_h,Ae_k)$ is satisfied for $A=0$. But we can't have weak convergence, since the principle of uniform boundedness implies that $\sup_n\lVert A_n\rVert$ is finite.
But we can show that the property you mentioned and $\sup_n\lVert A_n\rVert<\infty$ is equivalent to have weak convergence.
Indeed, if we have weak convergence then for all $v$, the sequence $\{A_n v\}$ converges weakly to $Av$, so it's bounded and $\sup_n\lVert A_n\rVert<\infty$.
Conversely, if $\sup_n\lVert A_n\rVert<\infty$ and $(e_h,A_ne_k)\to(e_h,Ae_k)$ for all $h,k\in\mathbb N$, we fix $u,v\in H$, and $\delta>0$. Let u',v'\in \operatorname{Span}\{e_h,h\in\mathbb N\} such that \lVert u-u'\rVert\leq \delta and \lVert v-v'\rVert\leq \delta. We have, writing $M:=\sup_{n\in\mathbb N}\lVert A_n\rVert$: \begin{align*} |(u,A_nv)-(u,Av)|&\leq |(u,A_nv)-(u',A_nv)|+|(u',A_nv)-(u',A_nv')|\\ &+|(u',A_nv')-(u',Av')|+(u',Av')-(u,Av')|+|(u,Av')-(u,Av)|\\ &\leq \lVert u-u'\Vert M\lVert v\rVert +\lVert u'\rVert M\lVert v-v'\rVert + |(u',A_nv')-(u',Av')|\\ &+\lVert u'-u\rVert \lVert A v'\rVert+ \lVert u\rVert\lVert A\rVert\lVert v-v'\rVert\\ &\leq \left(M\lVert v\rVert+M\lVert u'\rVert+\lVert Av'\rVert +\lVert u\rVert\lVert A\rVert\right)\delta+|(u',A_nv')-(u',Av')|, \end{align*} so for all $\delta >0$, $\limsup_n|(u,A_nv)-(u,Av)|\leq \left(M(\lVert v\rVert+\lVert u\rVert+\delta)+(\lVert v\rVert+\lVert u\rVert+\delta)\lVert A\rVert\right)\delta,$ hence $\lim_{n\to\infty}|(u,A_nv)-(u,Av)=0$ for all $u,v\in H$.