The easy part is to show that $f$ and $g$ have the same Fourier coefficients. The hard part is to show that $f$ and $g$ are identically equal.
First show that $f$ and $g$ have the same Fourier series.
From periodicity of the series, we can consider any interval $[\alpha, \alpha+ 2\pi]$, and adopting standard notation replace $a_0$ with $a_0/2$.
We have
$$\tag{1}g(x) = a_0/2 + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$$
where the series is uniformly convergent on the interval, and, since each term in the series is continuous, it follows from uniform convergence that $g$ is continuous.
Multiplying by $\sin mx$ and integrating we get
$$\tag{2}\frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}g(x) \sin mx \,dx \\= \frac{a_0}{2\pi}\int_{\alpha}^{\alpha + 2\pi}\sin mx \, dx +\frac{1}{\pi}\int_{\alpha}^{\alpha + 2\pi}\sum_{n=1}^\infty(a_n \cos nx \sin mx + b_n \sin nx \sin mx) \, dx.$$
Since (1) is uniformly convergent and $\sin mx$ is bounded, the series on the RHS of (2) is uniformly convergent and can be integrated term by term to obtain
$$\tag{3}\frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}g(x) \sin mx \,dx \\= \frac{a_0}{2\pi}\int_{\alpha}^{\alpha + 2\pi}\sin mx \, dx +\sum_{n=1}^\infty\frac{a_n}{\pi}\left(\int_{\alpha}^{\alpha + 2\pi}\cos nx \sin mx \, dx \right) + \frac{b_n}{\pi} \left(\int_{\alpha}^{\alpha + 2\pi} \sin nx \sin mx \, dx \right)$$
All the integrals on the RHS of (3) vanish except the integral of $\sin nx \sin mx$ when $n = m$.
Hence,
$$\frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}g(x) \sin mx \,dx = b_m = \frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}f(x) \sin mx \,dx $$.
Similarly we can show
$$\frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}g(x)\,dx = a_0 = \frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}f(x)\,dx , \\ \frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}g(x) \cos mx \,dx = a_m = \frac{1}{\pi} \int_{\alpha}^{\alpha + 2\pi}f(x) \cos mx \,dx. $$
Thus $f$ and $g$ have the same Fourier series.
It still remains to prove that $f = g$.
This follows by showing that if two continuous functions differ at just one point, $f(c) \neq g(c)$, then they cannot have the same Fourier series.
Take $h = f -g$. If $f$ and $g$ have identical Fourier series, then all Fourier coefficients of $h$ vanish, and for any trigonometric polynomial $T_m(x) = A_0/2 + \sum_{n=1}^{m} (A_n \cos nx + B_n \sin nx)$ we have for any $\alpha \in \mathbb{R}$,
$$\tag{4}\int_{\alpha}^{\alpha + 2\pi} h(x) T_m(x) \, dx = 0.$$
We also have $h(c) = f(c) - g(c) \neq 0$ and WLOG can assume that $h(c) > 0$. Since $h$ is continuous , there exists $K > 0$ and $\delta > 0$ such that $h(x) \geqslant K > 0$ when $x \in [c - \delta,c + \delta]$.
It can be shown that $T_m(x) = [1 + \cos(x-c) - \cos \delta]^m$ is a trigonometric polynomial satisfying
$$T_m(x) \geqslant 1 \text{ for } x \in [c-\delta,c+\delta] \\ \lim_{m \to \infty} T_m(x) = \infty \text{ uniformly on } [c - \delta/2, c+\delta/2] \\ |T_m(x)| \leqslant 1 \text{ for } x \in [c + \delta , c - \delta + 2\pi]
$$
From these properties it follows that
$$\tag{5} \int_{c - \delta}^{c + \delta} h(x) T_m(x) \, dx \geqslant \int_{c - \delta/2}^{c + \delta/2} h(x) T_m(x) \, dx \geqslant \delta \, K \, \inf T_m(x) $$
and $\int_{c + \delta}^{c - \delta + 2\pi} h(x) T_m(x) \, dx$ is bounded.
Since the RHS of (5) tends to $\infty$ as $m \to \infty$ it follows for sufficiently large $m$
$$\int_{c - \delta}^{c - \delta + 2\pi} h(x) T_m(x) \, dx \neq 0,$$
contradicting (4) and leading to the conclusion that if $f$ and $g$ differ at one point, they cannot have the same Fourier series.
Best Answer
$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\paren#1{\left(#1\right)}$The counter-example provided by @Conrad is enough to disprove the proposition. A stronger proposition, however, can be proved:
Proof: Denote the partial sums of the Fourier series by $S_N(x) = a_0 + \sum\limits_{n = 1}^N (a_n \cos nx + b_n \sin nx)$ for $N \in \N$. Note that the Dirichlet condition implies that$$ \lim_{N → ∞} S_N(x) = \frac{1}{2} (g(x + 0) + g(x - 0)).\quad \forall x \in \R $$ Suppose $\{S_N\}$ converges uniformly on $(0, π)$, then\begin{gather*} \frac{1}{2} (g(π + 0) + g(π - 0)) = \lim_{N → ∞} S_N(π) = \lim_{N → ∞} \lim_{x → π-} S_N(x)\\ \stackrel{(1)}{=} \lim_{x → π-} \lim_{N → ∞} S_N(x) = \lim_{x → π-} \frac{1}{2} (g(x + 0) + g(x - 0)) \stackrel{(2)}{=} g(π - 0), \end{gather*} where (1) uses uniform convergence and (2) uses monotonicity of $g$ on $(-π, π)$. But$$ g(π - 0) = f(π - 0) \geqslant f\paren{ \frac{π}{3} } > f\paren{ -\frac{π}{3} } \geqslant f(-π + 0) = g(-π + 0) = g(π + 0),$$ a contradiction.