Let $H$ be a separable Hilbert space, $\{|u_i\rangle\}_{i=1}^{\infty}$ be an ONB and $\hat{\mathcal{A}}\colon H\longrightarrow H$ be a continuous linear operator whose domain equals the whole Hilbert space.
I assumed that the expression $$ \sum_{j=1}^\infty\sum_{i=1}^\infty \mathcal{A}_{ij}|u_i\rangle\langle u_j|,\space\space\space \mathcal{A}_{ij} = \langle u_i|\hat{\mathcal{A}}|u_j\rangle. $$ should be valid without requiring $\hat{\mathcal{A}}$ to be Hilbert-Schmidt, because that sum would act on any $|v\rangle\in H$ like this: $$ \sum_{j=1}^\infty\sum_{i=1}^\infty \mathcal{A}_{ij}|u_i\rangle\langle u_j|v\rangle = \sum_{j=1}^\infty\sum_{i=1}^\infty \mathcal{A}_{ij}v_j|u_i\rangle = \hat{\mathcal{A}}|v\rangle \implies \hat{\mathcal{A}} = \sum_{j=1}^\infty\sum_{i=1}^\infty \mathcal{A}_{ij}|u_i\rangle\langle u_j|. $$ However, I keep reading that it is only well-defined for Hilbert-Schmidt operators. I think this is because the limit is taken with respect to the Hilbert-Schmidt norm, but can't the supremum norm be used to define that sum for any continuous operator?
Ah, yes, you are certainly correct that that infinite sum would give the operator... But/and the question is "where" does it converge, as a thing itself, rather than just giving correct outputs.
Well, yes, thinking in terms of the Schwartz Kernel Theorem (overkill here...), there is some sort of two-variable function that gives any operator from a Hilbert space to itself... in fact, much more generally.
The issue in making precise sense of such an infinite sum expressing (in the obvious way) this operator is about its convergence. Your observation that "it gives the right answer" is exactly an assertion that it converges in some sort of weak topology on operators... Bingo.
The objection about Hilbert-Schmidt operators is that that sum does not generally converge in the more obvious/elementary topology that tries to make a Hilbert space out of operators on a given Hilbert space. (Either by defining the Hilbert-Schmidt norm and taking completion of finite-rank operators, or by taking the subset of continuous operators with finite Hilbert-Schmidt norm...)
In summary, it's not so easy to justify that expression in relatively elementary terms, but it is completely fine if we use more sophisticated notions of convergence. :)