For a countably infinite dimensional vector space, the localisation of $A= End_k(V)$ at surjective endomorphisms is the zero ring If $k$ is a field and $V$ a countably infinite dimensional vector space, let $A = End_k(V)$ and $S = \{ s \in A : s \text{ is surjective} \}$. I am trying to show that $S^{-1}A$ is the zero ring. 
I have shown that $S$ is multiplicatively closed.  

It seems to me that $A$ can be considered as infinite matrices where the columns eventually have only zero entries.  

1.) Is this right?

Assuming this, it seems to me then that in much the same way for the finite dimensional case for a fixed $s \in S$, for any $a \in A$ we can find a $b \in A$ so that $sb=a$ 

2.) Is this true?

It then follows that for any $s^{-1}a \in S^{-1}A$, $s^{-1}a = b$ and further, there exists $s’ \in S$ such that $s’a = s$, so $s^{-1}a = s’^{-1}$. 

Now my formal definition of $S^{-1}A$ is:
$$\frac{A\langle i_s : s \in S \rangle}{\langle si_s - 1, 1 - i_ss \rangle} $$ 

And I have shown that any $i_sa = i_{s’}$ - a formal symbol, AND $i_sa = b\in A$. Which shows $S^{-1}A=0$ ?
The answers to your questions (1) and (2) are yes, but your subsequent arguments have some errors.  In particular, it is not true that there exists $s'\in S$ such that $s'a=s$ (consider what happens if if the kernel of $a$ is not contained in the kernel of $s$).  And in any case, nothing you have written would prove that $S^{-1}A=0$.  (Notably, as far as I can tell you have never used the assumption that $V$ is infinite-dimensional, so your argument, if it were valid, would work when $V$ is finite-dimensional.  But the conclusion is false in that case, since all surjections are already invertible $S^{-1}A=A$.)

Here is the direction you want to be thinking here.  Suppose you have a surjection $s\in S$ with kernel $W\subset V$ and $a\in A$ has image contained in $W$.  Then $sa=0$, so in $S^{-1}A$, $a$ becomes $0$.  So there are a lot of elements of $A$ that become $0$ in $S^{-1}A$.  If you could use this line of argument to eventually show that the unit element $1\in A$ becomes $0$ in $S^{-1}A$, then you could conclude that $S^{-1}A=0$.
Attempt number two following the very generous hint in previous answer: 

$E_{ii} \in A$ has image in $\{(0, ..., 0, a_i, 0, ...) : a_i \in k\}$.  

Then letting $s_i \in S$ be the map $(a_1, a_2, ....) \mapsto (a_{i+1}, a_{i+2}, ...)$ we see that $sE_{ii} = 0$ and so $E_{ii} = 0$ and hence $I = \sum_i E_{ii}$ is zero in $S^{-1}A$ and we’re done?