If by principal $\mathbf{G}_a$-bundle you mean a $\mathbf{G}_a$-torsor in the fppf topology, then this you can proved as follows. Since such objects are classified by elements in the fppf cohomology of your scheme with coefficients in $\mathbf{G}_a$, the result follows from the following proposition.

Proposition: Let $S$ be an affine scheme, and consider the group scheme $\mathbf{G}_{a}$ over $S$. Then $$H^1_{\text{fppf}}(S, \mathbf{G}_a) = 0.$$

To prove this, we first note by a theorem of Grothendieck from Brauer III that since $\mathbf{G}_a$ is smooth over $S$, we have $H^1_{\text{fppf}}(S, \mathbf{G}_a) = H^1_{\text{ét}}(S, \mathbf{G}_a)$. But now by Theorem 2.1 here on the comparison between Zariski and étale cohomology,
$$H^1_{\text{ét}}(S, \mathbf{G}_a) = H^1(S, \mathcal{O}_S)$$ which vanishes since $S$ is affine.

If by principal $\mathbf{G}_a$-bundle you mean a $\mathbf{G}_a$-torsor in the fppf topology, then this can proved as follows. Since such objects are classified by elements in the fppf cohomology of your scheme with coefficients in $\mathbf{G}_a$, the result follows from the following proposition.

Proposition: Let $S$ be an affine scheme. Then $$H^1_{\text{fppf}}(S, \mathbf{G}_a) = 0.$$

Proof: Since $\mathbf{G}_a$ is smooth (over $S$), by a theorem of Grothendieck from Brauer III we have $$H^1_{\text{fppf}}(S, \mathbf{G}_a) = H^1_{\text{ét}}(S, \mathbf{G}_a).$$ On the other hand, by Theorem 2.1 here on the comparison between Zariski and étale cohomology,
$$H^1_{\text{ét}}(S, \mathbf{G}_a) = H^1(S, \mathcal{O}_S).$$ The group on the right vanishes since $S$ is affine, proving the proposition.