Let $f:\mathbb{N}\to\mathbb{N}\cup\{0\}$ be a function. Let $f(1)=1,$ and $f(ab)=f(a)f(b)$ whenever $\gcd(a,b)=1.$ Note that I am assuming that $f$ is multiplicative but not completely multiplicative. Let $\mathbb{P}$ be the set of all primes. Assume there exists a $p_0\in\mathbb{P}$ such that $f(p_0)≠0.$ Further, let $f(p+q)=f(p)+f(q)$ whenever $p,q\in\mathbb{P}.$ This is what I mean by a function that is multiplicative, non-zero at a prime, and "prime-linear". Is it true that $f=\mathrm{id}_{\mathbb{N}}?$
I think $f$ is indeed the identity function. I do not have a full proof of this, but I've made some progress:
We begin by showing that $f(2)=2.$ Let $p_0$ (as given in the question) not be $2.$ Then, $f(2p_0)=f(2)f(p_0)$ since $\gcd(2,p_0)=1.$ Also, $f(2p_0)=f(p_0+p_0)=f(p_0)+f(p_0)=2f(p_0).$ So, $f(2)f(p_0)=2f(p_0).$ Since $f(p_0)≠0,$ we get that $f(2)=2.$ Suppose $p_0=2.$ Then, $f(10)=f(2)f(5)=f(5+5)=f(5)+f(5)=2f(5).$ Now, we show that $f(5)≠0.$ For the sake of contradiction, let $f(5)=0.$ Then, $f(5)=f(2)+f(3)=0.$ This gives $f(2)=-f(3).$ Since $f(2)\geq0,$ we get that $f(2)=0.$ This contradicts the fact that $f(p_0)=f(2)$ is unequal to $0.$ So, $f(5)≠0.$ Then, from $f(2)f(5)=2f(5),$ we can cancel out $f(5)$ and get $f(2)=2.$ From this, we also get $f(4)=4.$ Now, $f(12)=f(4)f(3)=4f(3)=f(7)+f(5)=f(3)+f(4)+f(2)+f(3)=6+2f(3).$ So, $4f(3)=6+2f(3).$ Hence, $f(3)=3.$ So far, we have: $$f(1)=1,$$ $$f(2)=2,$$ $$f(3)=3,$$ $$f(4)=f(2+2)=f(2)+f(2)=4,$$ $$f(5)=f(2+3)=f(2)+f(3)=5,$$ $$f(6)=f(2)f(3)=6,$$ $$f(7)=f(2)+f(5)=7,$$ $$f(8)=f(3)+f(5)=8,$$ $$f(9)=f(2)+f(7)=9,$$ $$f(10)=f(2)f(5)=10.$$ Now, I proceed by induction. I treat $1$ through $10$ as my base cases. Now, let $f(n)=n$ for each $n\in\{1,2,3,\ldots,k\}$ for some natural $k>10.$ Consider $f(k+1).$ If $k+1$ is composite and not a perfect prime power, then, $k+1=ab$ for some $a,b\in\{1,2,3,\ldots,k\},$ and $\gcd(a,b)=1.$ So, $f(k+1)=k+1.$ If $k+1$ is prime, it is odd since $k+1>2.$ So, $k$ is even. This implies that $k+4$ is even too. So, $k+4$ is composite. If $k+4$ is not a prime power, $f(k+4)=k+4$ since it can be written as $ab$ with $a,b\in\{1,2,3,\ldots,k\},$ and $\gcd(a,b)=1.$ Now, $f(k+4)=f(k+1+3)=f(k+1)+f(3)=f(k+1)+3=k+4.$ So, $f(k+1)=k+1.$ If $k+4$ is a prime power, it must be a power of $2$ since it's even. In that case, $k+6$ is not a prime power as it too, must be a power of $2$ if it is, and powers of $2$ cannot be seperated by $2$ unless they are $2$ and $4.$ So, $k+6=ab$ with $a,b\in\{1,2,3,\ldots,k\},$ and $\gcd(a,b)=1.$ Hence, $f(k+6)=k+6.$ Now, $f(k+6)=f(k+1+5)=f(k+1)+5=k+6.$ Hence, $f(k+1)=k+1.$ However, I am unable to deal with the case where $k+1$ is itself a prime power.
Note: As an interesting aside, if we assume Goldbach's conjecture, then, using the induction assumption, we may show that $f(2a)=2a$ for all $2a$ that can be written as a sum of primes $p$ and $q$ with $p,q\in\{1,2,3,\ldots,k\}.$
This answer is conditional on Goldbach's conjecture. If we assume it holds, we are almost done from what you have, and a lot of what you have generalizes. Suppose $f(n)=n$ for $n<k$. Then you've shown that $f(k)=k$ if $k$ is not a prime power. Now suppose $k$ is a prime power.
If $k$ is even (i.e. $k=2^m$), then by Goldbach's conjecture $k=p+q$ for some $p,q\in\mathbb{P}$ with $p,q<k$, so $f(k)=f(p+q)=f(p)+f(q)=p+q=k$.
If $k=r^m$ is odd, then $f(2k)=f(2)f(k)$, and also $f(2k)=f(p+q)=f(p)+f(q)$ with $p,q\in\mathbb{P}$ and $p,q<2k$ by Goldbach's conjecture. WLOG assume $3\le p<k$ and $k<q\le 2k-3$. Then $f(p)=p$ by the inductive hypothesis, so it suffices to show that $f(q)=q$. Similarly to how you showed it, we have that $f(q+3)=f(q)+f(3)$ since $3,q\in\mathbb{P}$. Also $q+3$ would be even, so $q+3=2^a\cdot b$ with $b$ odd. There are a few different cases here: $q+3=2k$, $b=1$, or neither. If neither $q+3=2k$ or $b=1$ holds, then $2^a,b<k$, so $f(q+3)=f(2^a)f(b)=2^ab$ by the inductive hypothesis.
If $b=1$, then $q+3$ is a power of $2$. This then means that $q+7$ can't be a prime power, but would be divisible by $4$ and not $8$ since $q > k\ge 10$. So $q+7=4c$ with $c$ odd. Then $f(4)f((q+7)/4)=f(q+7)=f(q)+f(7)$. Since $k>2$ and $q<2k-3$, we have that $\frac{q+7}{4}<k$. By the inductive hypothesis, this means that $f((q+7)/4)=(q+7)/4$. Then $4\cdot \frac{q+7}{4}=f(q)+7$, which means that $f(q)=q$.
Now suppose $q+3=2k$. Then $q+5=2k+2=2(k+1)$. Since $k$ is odd, this means that $q+5$ is divisible by $4$. Write $q+5=2^c\cdot d$ with $c\ge 2$ and $d$ odd. Then if $d\not=1$, then $2^c,d<k$, so $f(q)+f(5)=f(q+5)=f(2^c)f(d)=2^c\cdot d=q+5$, so $f(q)=q$.
The only case that remains to be considered is if $k$ is an odd prime power, $q+3=2k$ with $q$ prime, and $q+5$ is a perfect power of $2$. Then $q+13$ is a multiple of $8$, but not $16$ (as long as we loosen $k$ to be greater than $18$, which can be manually checked). Then $f(q+13)=f(q)+f(13)=f(8\cdot (q+13)/8)=f(8)f((q+13)/8)$. Since $f(13)=13$ and $(q+13)/8=(2k+10)/8 < k$, by the inductive hypothesis, this means that $f(q)+f(13)=q+13$, and since $f(13)=13$, this means that $f(q)=q$.
Thus, then $f(k)=f(p+q)=f(p)+f(q)=p+q=k$.
I think an answer independent of Goldbach's conjecture is possible, but I do think it would need a completely different approach.