Isomorphic: Properties Proved for A are True for B? To make this a little more concrete, consider vector spaces. 
An isomorphism between two spaces is an invertible linear transformation. It seems to then be commonly asserted or assumed that if A and B are isomorphic vector spaces they are to all intents and purposes the  same, and whatever   is proved for A is true for B. Is there a proof for this assertion ?
For example, if X is a basis in A and $\phi$ is an isomorphism $\phi$: A $\to$ B, then $\phi$(X) is a basis in B. I can write a specific proof for this, but could I  alternatively invoke the magic word isomorphism to declare it true ?

Edit after first answer:
If I prove that two vector spaces A and B are isomorphic, what of the theorems that I've proved about A can I assert about B ?
For example, if I prove that the intersection of two subspaces that make up an external direct sum is {0}, can I assert the same is true about an isomorphic internal direct sum or must I repeat the proof ?

Later Edit (2017)
I would hope to see some provable statement that a logical statement which is true about an object and only involves properties preserved by isomorphism (/homeomorphism) is equally true about an isomorphic object once the elements of the object are mapped by the implicit bijection.
 A: There can be no proof, because in this generality the assertion is false.
Basically, properties that hold for isomorphic objects are properties that only depend on the structure at hand, and not on the specific realization. There is a caveat, though: two objects may have more than one structure on them and be isomorphic with respect to one, but not the other.
For example, any two complex tori are diffeomorphic (i.e. isomorphic as differential manifolds), but there are infinitely many non-holomorphic (i.e. isomorphic as complex manifolds) complex tori.
Another example: $\Bbb{C}$ and $\Bbb{R}^2$ are isomorphic as real vector spaces, but they cannot be isomorphic as real algebras because their ring structures are different. This is apparent, because $\Bbb{C}$ is a field while $\Bbb{R}^2$ contains zero-divisors, like $(1,0)(0,1) = (0,0)$.

As for your second question: there is no hard and fast rule to decide which statement holds for isomorphic object and which doesn't.
In general, though, if your proof relies only on the structure with respect to which you are considering the isomorphism, then it is very likely that the same proof holds for the isomorphic object.
In your example, even a surjective morphism (of vector spaces) between $A$ and $B$ is enough: if $A = A_1 \oplus A_2$, $B = B_1 \oplus B_2$, and $\varphi \colon A \to B$ is a linear map such that $\varphi(A_1) = B_1$ and $\varphi(A_2) = B_2$, then by linearity $0 = \varphi(0) = \varphi(A_1 \cap A_2) = \varphi(A_1) \cap \varphi(A_2) = B_1 \cap B_2$.
On the other hand, you cannot say that every real algebra isomorphic to $\Bbb{C}$ as a real vector space is a field. This is because linear maps preserve the additive structure and the scalar product, but ignore the inner product.
As already suggested in the comments, you may wish to have a look at category theory, which formalizes mathematical structures in terms of morphisms. This means that whatever you can prove in the context of a given category about one of its objects is automatically true for every isomorphic object.
