How is "p implies q" same as "q unless not p"?

It doesn't!

"p implies q", $p\to q$ is equivalent to "q or not p", $q\vee \neg p$.   This is an inclusive or; it does not exclude the possibility that $q$ and $\neg p$ may be both true.

That is not the same thing as "q unless not p", which would be an exclusive or.

First, "unless" is not formal mathematical language. It is loose terminology.

Read it longer as "$q$ is true unless $p$ is false." This is read as "either $q$ is true or $p$ is not true."

If this is true, and $p$ is true, it in necessarily the case that $q$ is true.

Suppose: "If $p$, then $q$."

There are two cases: $p$ can be true, or it can be false. If $p$ is true, then so is $q$. So it follows that $q$ holds unless $p$ is false. Ergo, $q$ unless not $p$.

Now suppose: "$q$ unless $\neg p$."

We're trying to prove "If $p$, then $q$." So assume $p$. Since its not the case that $\neg p$ holds, hence from "$q$ unless $\neg p$" we deduce that $q$ holds. Ergo, if $p$, then $q$.