I thought that maybe the following product could be a good start to look for a metric in which consecutive primes are the same distance apart.
$$\frac{1}{\prod_{k=1}^\infty \big(p_k-1\big)^k}$$ where the product runs over the primes.
How do you construct a metric in which the primes are evenly spaced (1 unit). For example the natural numbers from one to the next always differ by exactly one unit. That is the gaps, are exactly $1$ unit from 2 to 3, from 3 to 4, from 4 to 5 and so on. How do you do this for the primes?
So the metric should be $d(p_n,p_{n+1})=1.$ Someone suggested using the discrete metric, in which case any two numbers would have distance equal to $1$ unit.
We define the function $\alpha:\Bbb N\to \Bbb R$ via $$ \alpha(k) = n + \frac{k-p_n}{p_{n+1}-p_n}, $$ where $n$ is such that $p_n\leq k < p_{n+1}$, with the exception $\alpha(1)=0$.
Note that this function is injective and satisifes $\alpha(p_n)=n$.
We then define our metric on $\Bbb N$ via $$ d(k,m):= |\alpha(k)-\alpha(m)|. $$ Using that $\alpha$ is injective and that $|\cdot|$ is a metric on $\Bbb R$, one can show that $d$ is indeed a metric on $\Bbb N$.
Moreover, we also have $$d(p_n,p_{n+1})=|\alpha(p_n)-\alpha(p_{n+1})| = |n-(n+1)|=1,$$ which satisfies your desired property.