If $P(n)$ has no connection with $P(n+1)$ but they are both true, can you call that "proof by induction"? For example in two sequences $a_1, a_2, a_3$... and $b_1, b_2, b_3$... I'd like to prove $P(n): a_n+b_n=b_n+a_n$. 
So I set the base case $P(1):a_1+b_1=b_1+a_1$, which is true by commutative law. 
Then I assume $P(n)$ is True. 
For $P(n+1)$ I use the commutative law, $P(n+1)$ is true. 
So $P(n)$ is true implies $P(n+1)$ is true. (?)
Can you call that "proof by induction"?
 A: For a proof by induction, you need to 1) prove the base case $P(1)$ (or whatever number the base case happens to be), 2) make the assumption for the inductive step $P(n)$ (strong or regular induction), and 3) then show that $P(n) \Rightarrow P(n + 1)$.
If you can do those three things, then you've completed a proof by induction. You aren't required to use any information from $P(n)$ when proving $P(n + 1)$ if you don't need to.

Intuitively, how does this apply to your case? Well, you showed the statement is true for $a_1$ and $b_1$. Because you did the inductive step, you can let $n = 1$ in $P(n)$ so that you know $P(2)$ is true. Then because you know $P(2)$ is true, you know $P(3)$ is true, and so on.

Another way you can think about it: look up the truth table for the logical implication. I'm referring specifically to $P(n) \Rightarrow P(n + 1)$. This is false only when $P(n)$ is true and $P(n + 1)$ is false. It is true at all other times.
Ask yourself this: is there ever a situation when $P(n)$ is true and yet $P(n + 1)$ is false? If the answer is no, then you're done.
A: Yes, when written in this form, the argument is a proof by mathematical induction, but it's unnecessarily so.  While you can wrap up the argument in the veneer of mathematical induction, in my personal opinion it would be disrespectful to any reader to leave it in that form.
After you've written a proof, you ought go through the proof and reduce it as close to its essence as you can.  If you can, simplify it as well.
In this case, you might as well say

Let $n$ be an arbitrary positive integer.  Then $P(n) : a_n + b_n = b_n + a_n$ is true by the commutative law.

If you don't need to reference $P$ in any later stage of the proof, this may be reduced further by

Let $n$ be an arbitrary positive integer.  Then $a_n + b_n = b_n + a_n$ is true by the commutative law.

