In $C^\ast$algebras we use $K(H)$, the algebra of compact operators on a separable Hilbert space, for stabilization of a $C^\ast$algebra, i.e. $S(A):=A\otimes K(H)$. Is there any similar stabilization functor in Banach algebras? What is the substitute of $K(H)$?

This is not an answer, but too long for a comment: Even for $C^*$algebras there is more than one method of stabilization. The critical feature of the compact operators is that $K(H) \otimes K(H) \cong K(H)$ for an infinite dimensional Hilbert space and that $K_*(K(H)) \cong K_*(\mathbb{C})$. But this also holds true for other $C^*$algebras like for example the infinite Cuntz algebra $\mathcal{O}_{\infty}$ or the JiangSu algebra $\mathcal{Z}$. In general  again for the $C^*$case  you might want to search for strongly selfabsorbing $C^*$algebras. There is the following theorem:
And there is a famous theorem by Kirchberg, which you may read like this:
If you are looking for a setup like this, you may want to search for Banach algebras, which have the $K$theory of $\mathbb{C}$ and are isomorphic to the tensor product of two copies of themselves. Or one might be able to generalize the notion of strongly selfabsorbing to Banach algebras. That was a lot about $C^*$algebras and not much about the Banach case. Sorry about that. 


There is now a theory of $L^p$operator algebras, defined mainly by Chris Phillips. (See for example http://arxiv.org/pdf/1309.6406.pdf.) The objects there considered are normclosed subalgebras of $B(L^p(X,\mu))$ for $\sigma$unital spaces $(X,\mu)$. (Note that even when $p=2$, these are not necessarily $C^*$algebras.) Of course not every Banach algebra has this form, but they are a considerable generalization of C*algebras (or operator algebras acting on Hilbert spaces), and one can use them as a test case for what could be a general theory for Banach algebras. In this context, what seems to be the right analog of stabilization (tensoring with $\mathcal{K}(\mathcal{H})$ in the C*algebra case), is tensoring with the closure of the union of the finite matrices. When $p=2$, what we get is just the compact operators, and this is also the case whenever $p>1$, but for $p=1$ one gets something smaller (which does not contain all rank one projections). One reason why this seems to be the right analog is that Takai duality seems to hold with this algebra in place of the usual $\mathcal{K}(\mathcal{H})$. In general, tensor products of Banach algebras are not so well behaved. One takes the algebraic tensor product first, and then one should complete in a suitable norm. Even in the $C^*$algebra case there is more than one choice, but the situation is much worse even in the $L^p$operator algebra case. Alain already mentioned the ambiguity of tensor products in $C^*$algebras, which is not an issue if one of the algebras is nuclear (=amenable). For $L^p$operator algebras, it is not known whether amenability is equivalent to nuclearity, and even in the presence of both, defining a tensor product is a complicated issue. I don't know what applications you had in mind, but you can start by looking at $L^p$operator algebras and tensor them with the closure of the union of the finite matrices, and see whether the statements you were hoping would be true, actually hold in this case. 

