Proof verification: AB5) implies AB4) in Grothendieck's Tohoku paper

Following the proof outlined in here, I wanted to show that when

\begin{alignedat}{3} 0 \longrightarrow \mathord{} & K_j & \mathord{} \longrightarrow \mathord{} & A_j & \mathord{} \longrightarrow \mathord{} & B_j \\ & \downarrow && \downarrow && \downarrow \\ 0 \longrightarrow \mathord{} & K & \mathord{} \longrightarrow \mathord{} & A & \mathord{} \longrightarrow \mathord{} & B \\ \end{alignedat}

commutes, the rightmost arrow being mono and both rows being exact, then the left square is a pullback. However, I only needed the bottom row to be exact at $$A$$, which threw me off a bit:

If you have $$\require{AMScd} \begin{CD} U@>>>A_j\\ @VVV @VVV\\ K @>>> A \end{CD}$$

Then by exactness at $$A$$ and commutativity, you have $$\begin{CD} U @>>> A_j @>>> B_j\\ @.@.@VVV\\ @.@.B \end{CD}$$ equal to $$0$$, and then because $$B_j \longrightarrow B$$ is mono, $$U \longrightarrow A_j$$ factors uniquely through $$K_j \longrightarrow A_j$$. The other leg of the pullback diagram commutes by uniqueness of the kernel factorisation through $$K \longrightarrow A$$.

Does that proof seem correct?

You have the right idea, but you really need the bottom row to be exact at $$K$$ and not necessarily at $$A$$.
Indeed, to get the factorisation of $$U\to A_j$$ through $$K_j$$ you need to prove that the composition $$U\to A_j\to B_j$$ is zero, and for this you only need to know that $$K\to A\to B$$ is zero, which is weaker than exactness at $$A$$.
On the other hand, to prove that the second leg of the pullback commutes, you use the uniqueness of the factorisation through $$K$$, or in other words, the fact that $$K\to A$$ is a monomorphism, which is equivalent to the exactness of $$0\to K\to A$$ at $$K$$.