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With one exception, the following proof is solely based on vector space axioms. Axiom names are italicised. They are defined in Wikipedia (vector space).
Vector spaces - If an addend adds nothing, then the addend is the zero vector.
Let $V$ be a vector space. By this proof, we know that a zero vector of $V$ is unique; let $0$ be the zero vector of $V$. Let $v_1, v_2 \in V$.
If $v_1 + v_2 = v_1$, then $v_2 = 0$.
Proof. We assume that $v_1 + v_2 = v_1$. It remains to prove that $v_2 = 0$. Let $(-v_1)$ be an additive inverse of $v_1$. \begin{align*} 0 &= v_1 + (-v_1) && \text{by }\textit{Inverse elements of addition} \\ &= (v_1 + v_2) + (-v_1) && \text{by assumption} \\ &= (v_2 + v_1) + (-v_1) && \text{by }\textit{Commutativity of addition} \\ &= v_2 + (v_1 + (-v_1)) && \text{by }\textit{Associativity of addition} \\ &= v_2 + 0 && \text{by }\textit{Inverse elements of addition} \\ &= v_2 && \text{by }\textit{Identity element of addition} \\ \end{align*} QED
The proof is correct. More intuitively, the proof is by cancelling $\,v_1$ from both sides, by adding its inverse, which makes it clear that the proof generalizes to the following cancellation law
$$ v+ u = v + w\ \Rightarrow\ u = w $$
This simple inference, that invertible elements are cancellable, is ubiquitous in algebra.