Let $A$ be a unital C*-algebra and denote by $F(H)$ the algebraic ideal of finite rank operators on $H = \ell^2(\mathbb N)$. Is is true that $$ A \odot F(H) $$ is an algebraic ideal inside $A \otimes \mathbb K$ ? Here $A \odot F(H)$ denotes the algebraic tensor product of $A$ and $F(H)$.
I have the feeling that this is not true, since it might happen (?) that that the product $(a \otimes f)x$ with $a \in A, \ f \in F(H)$ and $x \in A \otimes \mathbb K$ is not an element of the algebraic tensor product but rather an element of the closure.
I think your intuition is correct, even in the case of Abelian $C^\ast$-algebras. Take $A=c_0(\mathcal{Z})$. We have $$ A \otimes K(H) \simeq c_0(\mathbb{N}; K(H)). $$ For any element $x=(x_i)_i \in c_0(\mathbb{N}; K(H))$ we can define the following quantity $$ \mathrm{Im}(x) = \overline{\mathrm{span}}\big\{ \bigcup_{i} \mathrm{Range}(x_i) \big\} \leq H. $$ That quantity always gives a finite dimensional subspace if $x \in A \odot F(H)$. It only rests to see that there are cases in which $\mathrm{Im}(x \, f)$ is infinite, with $f \in A \odot F(H)$ and $x \in A \otimes K(H)$. Choose an ONB base $(e_i)_i$ of $H$ and denote by $(e_{i,j})$ the matrix units of $B(H)$. If $f_i = (i+1)^{-1} \otimes e_{1,1}$ and $x_i = (i+1)^{-1} \otimes e_{i,1}$, then $\dim \mathrm{Im}(x \, f) = \infty$.
Related: You can get a well-behaved algebraic ideal using the language of Hilbert $C^\ast$-modules. Take $A \odot H$ with the product given by $\langle a \otimes \xi, b \otimes \eta \rangle_X = a^\ast b \langle \xi, \eta \rangle$. And your ideal will be spanned by operators $\theta_{y,z}(x) = \langle y, x \rangle z$, see [Lance; p9].
Lance, E. Christopher, Hilbert $C^\ast$-modules. A toolkit for operator algebraists, London Mathematical Society Lecture Note Series. 210. Cambridge: Univ. Press,. ix, 130 p. (1995). ZBL0822.46080.