My suspicion is "no", because if I recall correctly the map $I \to V \otimes V^*$ naturally lands in the *injective tensor product*, not the *projective tensor product*, and it is the latter which appears as the ``correct'' tensor product for the SMC category of Banach spaces and linear contractions.

In the toy example given, $V\oplus V$ with the sup norm is the same as continuous maps from a 2-point set to $V$, equipped with sup-norm, and I'm pretty sure that this is indeed isometrically linearly isomorphic to ${\mathbb R}^2 \check{\otimes} V$ i.e. the injective tensor product.

EDIT: as Reid points out my remarks above assume without justification that the inj. t.p. does differ from the proj t.p. in the specific case being considered. I *think* this is indeed the case. Take $V$ to be ${\mathbb R}^2$ with usual Euclidean norm. The projective tensor product of $V$ with $V^\*$ can be identified with $M_2({\mathbb R})$ equipped with the trace class norm; the injective tensor product would lead to the `same' underlying vector space, equipped with the operator norm. The 2 x 2 identity matrix has trace class norm 2 and operator norm 1, so the two norms are genuinely different.

My answer is still not as clear as it should be, because due to a sluggish and temperamental internet connection I'm having trouble looking up just what the axioms for categorical duals in a SMC are. But if I recall correctly the natural map from $I \to V \otimes V^\*$ should be given by multiplying a scalar by the vector $e_1\otimes e_1 + e_2\otimes e_2$ where $e_1,e_2$ is an o.n. basis of ${\mathbb R}^2$ -- and that vector does not have norm 1 in the proj t.p. althought it does have norm 1 in the inj t.p.