The identity element of the Tensor Product of Vector Bundles I want to prove the following:
If $E$ is a smooth vector bundle over $M$, then $(M\times\mathbb{R})\otimes E \simeq E$. 
I built an isomorphism between the fibers, but I failed to define an isomorphism between the vector bundles.
 A: An isomorphism of vector bundles is just a morphism of vector bundles that is an isomorphism on every fiber (as pointed out by san in a comment).
You say you have an isomorphism between the fibers. I'll assume that the isomorphism you have is $\mathbb{R}\otimes F \to F : r\otimes f\mapsto rf$ where $F$ is a fiber.
Its inverse is $F\to\mathbb{R}\otimes F: f\mapsto 1\otimes f$.
The bundle isomorphism between $E$ and $(M\times\mathbb{R})\otimes E$ is obtained by taking the above isomorphism on every fiber:
$$ (M\times\mathbb{R})\otimes E \to E : (m,r)\otimes f_m \mapsto (rf)_m $$
where I used subscripts to indicate basepoints. The addition of a basepoint in the notation is really the only difference with the vector space case.
The inverse is
$$ E \otimes (M\times\mathbb{R})\otimes E : f_m \mapsto (m,1)\otimes f_m .$$
This answers your question (I hope).

(Optional extra paragraph.)
Note that finding the isomorphism of vector bundles was not actually harder than finding the isomorphism between fibers. We just needed to include a basepoint in the notation. This is often true: if you can find a natural map between vector spaces, you get the same map between vector bundles. For example, if $V$ and $W$ are vector spaces then $\text{Hom}(V,W) \cong V^*\otimes W$ by some natural map.
Using the same map but adding basepoints, we get that $$\text{Hom}(D,E) \cong D^*\otimes E$$ where $D$ and $E$ are vector bundles.
(Here by $\text{Hom}(D,E)$ I mean the vector bundle whose fiber over $m$ is $\text{Hom}(D_m,E_m)$. I don't mean the set of vector bundle morphisms from $D$ to $E$.)
