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.
We want to show that a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 1 if and only if a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 2.
Suppose that $T_{\lambda}\to T$ with respect to $\textbf{def}$ 1, then we know that $T_{\lambda}x\to Tx$ for each $x\in X$ meaning that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T).$$
Putting this in context, for each $F_x\in \{F_x:\beta(X,Y)\to Y \text{ such that }F_x(T)=Tx, T\in\beta(X,Y)\}$ we have $F_x(T_{\lambda})\to F_x(T)$. So $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2.
On the other hand, if $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2, then we know that
$$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T),$$
for each $x\in X$ and as a result $T_{\lambda}x\to Tx$ for each $x\in X$ and we have convergence with $\textbf{def}$ 1. It is simply unpacking the definitions.
Best Answer
The answer is in part I of Linear Operators by Dunford and Schwartz, Theorem VI.1.4. For any Banach spaces $X$ and $Y$ linear functionals on $\mathcal{B}(X,Y)$ continuous in strong and weak operator topologies are the same. It follows from the proof that they are of the form $F(A)=\sum_{i=1}^n\varphi_i(Ax_i)$ for $x_i\in X$ and $\varphi_i\in Y^*$. In other words, the dual space can be identified with the algebraic tensor product $X\otimes Y^*$, or equivalently the space of finite rank operators $\mathcal{F}(Y,X)$.
For the ultraweak and the ultrastrong topologies on $\mathcal{B}(H)$, where $H$ is a Hilbert space, the answer is in Von Neumann Algebras by Dixmier, Lemma I.3.2. Again, the dual spaces coincide and can be identified with the projective tensor product $H\otimes_\pi H$, or the space of the trace class operators $\mathcal{L}_1(H)$. It also happens to coincide with the norm closure of $\mathcal{F}(H)$ in the Banach dual $\mathcal{B}(H)^*$, and is a Banach predual $\mathcal{B}(H)_*$ of $\mathcal{B}(H)$. Moreover, the ultraweak topology on $\mathcal{B}(H)$ coincides with the weak* topology generated by this predual, they are both $\sigma\big(\mathcal{B}(H),H\otimes_\pi H\big)$.
It doesn't appear that the ultraweak or the ultrastrong topologies were studied much beyond Hilbert spaces, I couldn't even find any established definitions for them. Since the weak operator topology coincides with $\sigma\big(\mathcal{B}(X,Y), X\otimes Y^*\big)$, by analogy to Hilbert spaces a natural candidate for the ultraweak is $\sigma\big(\mathcal{B}(X,Y), X\otimes_\pi Y^*\big)$. However, according to Introduction to Tensor Products of Banach Spaces by Ryan, Section 2.2 a predual to $\mathcal{B}(X,Y)$ is $X\otimes_\pi Y_*$, where $Y_*$ is a predual to $Y$. So the ultraweak so defined does not coincide with the weak* unless $Y$ is reflexive. And even if $Y$ is reflexive it is not clear what a natural candidate for the ultrastrong is or how the duals are related.