Yes, that is a fine proof.
As a general principle, if you can write down an isomorphism of vector spaces without making any choices, then you have written down an isomorphism of vector bundles.
Let's make that principle precise in this case. On a small open set $U$, trivialize the vector bundles $V_1$ and $V_2$ with frames (:=sections which form a basis at every point-- note that, seemingly contra the principle, we've made a choice) $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ (I changed your notation $r_1, r_2$ to $m, n$ to avoid double subscripts).
Then the bundle $V_1 \otimes V_2$ is trivialized with frame $e_i \otimes f_j$ and its determinant trivialized with the singleton frame (alphabetical order)
$$
(e_1 \otimes f_1) \wedge (e_1 \otimes f_2) \wedge \ldots \wedge (e_1 \otimes f_m) \wedge(e_2\otimes f_1) \wedge \ldots \wedge (e_n \otimes f_m)
$$
which we may map to
$$
(e_1\wedge \ldots \wedge e_n)^m \otimes (f_1 \wedge \ldots f_m)^n.
$$
Now it seems so far that our map depends on choices, but it does not. We just need to check that multiplying an $e_i$ by an invertible function, adding $\phi e_i$ to $e_j$ (for $\phi$ an arbitrary function), or flip-flopping $e_i$ and $e_j$ does nothing (and same for $f_i$ and $f_j$). Note that in each case, the given bases for $\text{det}(V_1 \otimes V_2)$ and $\text{det} (V_1)^m \otimes \text{det} (V_2)^n$ multiply by the same scalar function, so the map doesn't change.
We were (allegedly) doing all the above reasoning on a small open set $U$ -- otherwise, there may not exist a frame (the existence of a frame on an open set being equivalent to a vector bundle being trivial). Now suppose we define a global map by doing the same reasoning on ALL open sets. We have to check that if $U$ and $W$ are different, we have defined the same map on $U \cap W$.
But it follows from the independence of choices. The restriction of a choice of frame over $U$ to a frame over $U \cap W$ gives the map of bundles over $U \cap W$, and so does the restriction of a choice of frame over $W$. But we know that the map doesn't depend on a choice. So we've therefore given a global map of vector bundles.
In general it can be general to try to work even more abstractly, i.e. not in terms of picking bases, so that it becomes completely automatic, by the above principle, that a map of vector spaces extends to a map of vector bundles.
For any rank 2 vector bundle there is a pairing
$$
E \otimes E \to \det(E).
$$
Being non-degenerate it induces an isomorphism $E \cong E^\vee \otimes \det(E)$, which thus holds for any rank 2 vector bundle.
For the more general question --- such an isomorphism exists if and only if there is a non-degenerate pairing $E \otimes E \to L$.
Best Answer
No. Assume, for instance, that $L$ is a line bundle of order 2, i.e., $$ L^2 \cong \mathcal{O}, \qquad L \not\cong \mathcal{O}. $$ Let $E = \mathcal{O} \oplus L$. Then $$ E \otimes \mathcal{O} \cong E \cong E \otimes L, $$ but $L \not\cong \mathcal{O}$.