No, $f$ is not surjective, as the range is the set of sequences with only finitely many non-zero entries. For example, you will never have
$$f(n)=(1,2,3,4,\ldots)\ .$$
I will write rows with $[ ~~]$ brackets. (If you also admit an empty row $\emptyset$ you can just extend the bollow definition with $f(\emptyset) := 1$).
Then we can define a bijection $f : \mathbb{S} \rightarrow \mathbb{N}_{>1}$ by
\begin{align}
f([a_1, \dots, a_k]) := p_1^{a_1} \dots p_{k-1}^{a_{k-1}} \cdot p_k^{1 + a_k}
\end{align}
Where $p_k$ is the $k$-th prime. It is then easy to get a bijection $\mathbb{S} \rightarrow \mathbb{N}$ by combining $f$ with a bijection $\mathbb{N}_{>1} \to \mathbb{N}$.
Proof of Surjectivity:
Any $n > 1$ has a biggest ($k$-th) prime number $p_k$ with exponent $e > 0$ s.t. $n = x \cdot p_k^e$ and $p_k \not \mid x$.
If $x = 1$ we have
$
f ([\underbrace{0, \dots ,0}_{k-1 \text{ times}},e-1] ) = p_k^{1 + (e-1)} = x \cdot p_k^e = n
$
If $x > 1$ then we have $x < n$, and therefore by complete induction $x$ has a preimage $[x_1, \dots, x_l]$. Then we have
\begin{align}
f ( [ x_1 , \dots , 1 + x_l, ~\color{Blue}{0, \dots ,0}, ~e-1 ] )
= p_1^{x_1} \dots p_l^{1 + x_l} \cdot p_k^{1 + (e-1)} = x \cdot p_k^e = n
\end{align}
Where we need to put $\color{blue}{\text{$(k-l)-1$ zeros}}$ between $(1 + x_l)$ and $(e-1)$ to make sure the latter is the $k$-th entry of the tuple.
Note that we must have $l < k$ since the biggest prime in $n$ was $p_k$ and all primes $p_l$ appearing in $x$ must be smaller. $\Box$
Proof of Injectivity:
Assume $f([a_1, \dots, a_n]) = f([b_1, \dots, b_m])$, then
$$
p_1^{a_1} \dots p_n^{1 + a_n} = p_1^{b_1} \dots p_m^{1 + b_m}
$$
where both sides are prime factorisations of the same number. Since $\color{Red}{\text{said number is bigger then } 1}$, we can conclude $n = m$ and $a_j = b_j$ by the uniqueness of prime factorizations. $\Box$
Note that it is in the proof for injectivity, that the (seemingly odd) $+1$ in the power of the last prime number $\color{Red}{\text{plays its crucial role}}$. Without it, the proof would not work.
Best Answer
Xavier shows a "non-constructive" proof of the fact that $\#\mathbb P=\#\mathbb N$. I will show a constructive one:
Let $F_n=2^{2^n}+1$ be the $n$-th Fermat number. It is proved that Fermat numbers a coprime. Let $P_n$ be the smallest prime number dividing $F_n$. Then $P_n\neq P_m$ if $n\neq m$ (because the Fermat numbers are coprime hence having different primes in thei factorizations).
This means that the map $\mathbb N\to\mathbb P$ given by $n\to P_n$ is injective.
As well, the identity map $\mathbb P\to\mathbb N$ given by $p\to p$ is injective.
Therefore by Cantor-Bernstein theorem, we have $\#\mathbb P=\#\mathbb N$.