The statement "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" just means remove the weights occurring in the irreducible representation from those appearing in your representation, and see what's left. In your case, the irreducible representation with highest weight (1,1) is 8-dimensional (and is actually isomorphic to the adjoint representation), and its weights are listed in Section 6.5 of Hall: they are (1,1), (-1,2), (2,-1), (0,0) (twice), (1,-2), (-2,1), and (-1,-1). Removing them from your list of weights leaves only (0,0), which is the weight of the 1-dimensional trivial representation (with highest weight (0,0)). So your tensor product representation decomposes into the 8-dimensional adjoint representation plus a 1-dimensional trivial representation. This is sometimes written as $3\otimes\overline{3} = 8\oplus 1$.
In fact, if $L$ is any Lie algebra over any field $k$, and $L_1, L_2$ are subalgebras such that $L = L_1 \oplus L_2$ as $k$-vector spaces, then the $k$-linear map $f$ given by
$$U(L_1) \otimes_k U(L_2) \rightarrow U(L)$$
$$x \otimes y \mapsto xy$$
is an isomorphism of $k$-vector spaces (and hence of $(U(L_1),U(L_2))$-bimodules).
As noticed in the first bullet point in this question, this follows from the Poincaré-Birkhoff-Witt theorem, compare e.g. Bourbaki, Lie Groups and Algebras ch. 1 §7. Indeed, it is a corollary of PBW (in Bourbaki, Corollary 3) that if $(e_1, ..., e_n)$ is an ordered $k$-basis of a Lie algebra $\mathfrak g$, then the monomials $e_1^{r_1} \cdot ... \cdot e_n^{r_n}$ are a $k$-basis of $U(\mathfrak g)$. From this the assertion (which is stated in greater generality in Bourbaki's Corollary 6) follows. Namely, if now $(x_1, ..., x_m)$ is an ordered $k$-basis of $L_1$, and $y_1, ..., y_n$ one of $L_2$, then on the one hand PBW and tensor products over fields tell us that the set of all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \otimes y_1^{l_1} \cdot ... \cdot y_n^{l_n}$ is a $k$-basis of $U(L_1) \otimes U(L_2)$, on the other hand the images of these elements under $f$, namely, all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \cdot y_1^{l_1} \cdot ... \cdot y_n^{l_n}$, again by PBW are a $k$-basis of $U(L)$, because by assumption $(x_1, ..., x_m, y_1, ..., y_m)$ is an ordered $k$-basis of $L$.
Now apply this to $L=\mathfrak g, L_1 = \mathfrak{n}^-, L_2= \mathfrak b$.
To see where your supposed counterargument goes wrong (copied from comments and discussion):
Your computation of $xy=0$ for certain $x, y \neq 0$ seems to be happening in $M_3(\mathfrak k)$ (or $End_{\mathfrak k} ({\mathfrak k}^3)$). That matrix ring is an enveloping associative algebra of $\mathfrak{sl}_3$, and maybe likewise would be certain associative subalgebras $B:= \pmatrix{*&*&*\\0&*&*\\0&0&*}$, $N^-:=\pmatrix{0&0&0\\*&0&0\\*&*&0}$ corresponding to $\mathfrak b, \mathfrak n^-$. These algebras are quotients of the respective universal enveloping algebras (by the latter's universal property, although surjectivity might be more subtle), but of course a product being zero in a quotient tells us little about the product being zero in the original ring.
What this computation shows, then, is that even though there are canonical surjections of associative algebras
$$U(\mathfrak n^-) \twoheadrightarrow N^-, U(\mathfrak b) \twoheadrightarrow B, U(\mathfrak g) \twoheadrightarrow M_3(k)$$
the map $$N^- \otimes_{\mathfrak k} B \rightarrow M_3(k)$$
induced via those projections from the given map $U(n^-) \otimes U(b) \simeq U(g)$, is no longer an isomorphism. But viewed this way, that is maybe not as surprising.
Full disclosure, I have made the mistake of mixing up standard enveloping matrix algebras with the universal enveloping algebra before, and I still sometimes do calculations in those matrix rings to get a first feeling for the universal enveloping one, but it's important to remember that the matrix algebras (like the above $M_3(\mathfrak k), B, N^-$) are just quotients -- and for that matter, with a lot factored out -- of the "much, much bigger" universal enveloping algebras. In particular, note that the above PBW corollaries also imply that $x^my^n \neq 0$ in $U(\mathfrak g)$ for all $m,n \in \mathbb N$ and any $x \neq 0 \neq y \in \mathfrak g$.
Best Answer
First of all, it is unclear to me why you compute $\mathfrak{g}^{\otimes j}$. The universal enveloping algebra is a quotient of the tensor algebra, so $T^j(\mathfrak{g}) = \mathfrak{g}^{\otimes j}$ will not share the irreducible decomposition of $\mathcal{U}(\mathfrak{g})$.
Secondly, the decomposition of $\mathfrak{su}(2)$ seems wrong: recall that $\mathcal{U}(\mathfrak{g}) \cong \mathcal{S}(\mathfrak{g})$ (the symmetric algebra, which can be interpreted as polynomials in $\mathfrak{g}$) as $\mathfrak{g}$-modules. Then clearly $\mathcal{S}^2(\mathfrak{g})$ has dimension 6, generated by polynomials $e^2, f^2, h^2, eh, fe, fh$. At the same time, $e^2$ is a maximal vector of weight 4, so it generates an irreducible $\mathfrak{g}$-module isomorphic to $V_5$. It follows that the decomposition of $\mathcal{U}(\mathfrak{g})$ is $V_1 \oplus V_5$. But we also have $\mathcal{U}^0(\mathfrak{g}) \cong V_1$ (ie the scalars). Thus we see that $V_1$ appears multiple times in the decomposition, and not just once.
So let's think this through from the start. First of all, what is the decomposition of $\mathfrak{g} = \mathfrak{su}(2)$ as a $\mathfrak{g}$-module into simple modules? Well, we know $\mathfrak{su}(n)$ is simple, and a submodule of the adjoint representation is the same as an ideal. Therefore, the adjoint representation must be simple. For $\mathfrak{su}(2)$, we have simply $\mathfrak{su}(2) \cong V_3$ as $\mathfrak{su}(2)$-modules. Next, we need a way to compute the decomposition of symmetric powers of $V_3$. This sort of problem (with tensor, symmetrix or exterior powers) is called $\textbf{plethysm}$. It is a hard problem in general but many specific cases have been worked out. For the tensor product case, there is Steinberg's formula. The symmetric power case of $\mathfrak{su}(2)$ it is not too bad, see for instance this post.
For $\mathfrak{su}(3)$, we can see that $\mathfrak{su}(3) \cong L(\alpha_1 + \alpha_2)$. This is the irreducible module with highest weight vector having weight $\alpha_1 + \alpha_2$. Next we would need to find the decomposition of $\mathcal{S}^j(L(\alpha_1 + \alpha_2))$. Unfortunately I didn't find anything that would help with this through a cursory online search.