"Weak convergence of measures" is a misnomer. What it really means is that the space of measures is identified, via Riesz representation, with the dual of some space of continuous functions, and this gives us weak* topology on the space of measures. People just don't like saying that their measures converge weak-star-ly, or putting a lot of asterisks in their texts.
Folland writes in Real Analysis, page 223:
The weak* topology on $M(X)=C_0(X)^*$ ... is of considerable importance in applications; we shall call it the vague topology on $M(X)$. (The term "vague" is common in probability theory and has the advantage of forming an adverb more gracefully than "weak*".) The vague topology is sometimes called the weak topology, but this terminology conflicts with ours, since $C_0(X)$ is rarely reflexive.
I think what you are saying is true. Never thought about it since i've always pre-assumed that the weakly-operator limit $A$ of the $A_n's$ was always in $A\in \mathfrak L(X,Y)$. Am writing the argument just to convience ourselfs. Indeed, we only need to assume that $Y$ has a norm, not necessarily a complete one.
So, lets suppose that $A_n\overset{\text{wo}}{\to}A$ in the weak operator topology where $A:X\to Y$ is a linear operator, not necessarily bounded. Convergence in the weak operator topology is described by $h(A_n x)\to h(A x)$ for every $x\in X$ and $h\in Y^*$. This implies that the set $\{A_n x: n\in \mathbb{N}\}$ is weakly bounded in $Y$, hence it is also bounded in $Y$. By the Banach-Steinhaus it follows that $\sup_{n}||A_n||=M<\infty$. Now, for $x\in X$ with $||x||=1$ we have
$$||Ax||=\max_{h\in Y^*,\, ||h||=1}|h(Ax)|$$
So, there is some $||h||=1$ in $Y^*$ such that $||Ax||=|h(Ax)|$. Using the weak convergence for $A_nx$ we end up with
\begin{align}
||Ax||&=|h(Ax)|\\
&=\lim_{n\to \infty}|h(A_nx)|\\
&\leq \underbrace{||h||}_{=1}\liminf_{n\to \infty}||A_n||\cdot \underbrace{||x||}_{=1}
\end{align}
Hence, $||Ax||\leq M$ for every $||x||=1$ and therefore, $||A||\leq M<\infty$.
Edit: (Responding to the comment)
The existence of such $A$ is trickier. To ensure such existence we need another assumption for $Y$, since there is a counter example in here where $X=Y=c_0$. The only natural that i could think while i was trying to prove it is that $Y$ has to be reflexive (from not being a Banach space we went straight out to reflexivity :P). In the case where $X=Y=H$ is a Hilbert space things were slightly more easier since we can identify $H^*$ with $H$ and dont need to mess with the second duals.
The argument in the case where $Y$ is reflexive is the following:
Suppose that $\lim_{n}\langle A_n x, h \rangle$ exists for every $x\in X$ and $h\in Y^*$. For fixed $x\in X$ let $f_x:Y^*\to \mathbb{R}$ defined by
$$\langle h, f_x\rangle =\lim_{n\to \infty}\langle A_n x, h\rangle$$
Its easy to check that $f_x$ is a linear functional and by the previous discussion it is also bounded. Meaning, $f_x \in Y^{**}$. By reflexivity, there is some $y_x\in Y$ such that $\langle h, f_x\rangle =\langle y_x, h\rangle$ for all $h\in Y^*$. Now, let $x\overset{A}{\longmapsto} y_x$. Now, its easy to check that $A:X\to Y$ is a linear operator. By the previous discussion it is also bounded.
Best Answer
They are not equivalent on an infinite-dimensional Hilbert space. The weak convergence for operators in this case is in the usually called "weak operator topology": $$ A_n\xrightarrow{wot} A\ \ \iff\ \ \langle A_nx,y\rangle\to\langle Ax,y\rangle,\ \ \forall x,y\in H. $$ The weak operator topology is known to be coarser than the $\sigma$-weak operator topology, which is the one given by the normal functionals (which, among other characterizations, are precisely those wot-continuous on bounded sets). The weak topology is even finer, as it is given by all the functionals, including the non-normal ones (for instance, those that are zero on the compacts).