The use of the MVT given in the OP to justify the result is not correct since $c$ depends on the endpoint $x$.
Under the assumption that $\phi$ is strictly increasing and that $\phi'\neq0$ in $(a, b)$, it is possible to prove the result using the inverse function theorem. For simplicity, assume $\phi'>0$ on $(a, b)$. Then
\begin{align}
\int^b_a f(t) e^{ix\phi(t)}\,dt&=\int^b_a f(\phi^{-1}(\phi(t))\frac{\phi'(t)}{\phi'(\phi^{-1}(\phi(t)))}e^{ix\phi(t)}\,dt\\
&=\int^{\phi(b)}_{\phi(a)}f(\phi^{-1}(u))\frac{1}{\phi'(\phi^{-1}(u))}e^{ixu}\,du\\
&=\int^{\phi(b)}_{\phi(a)} F(u)\,e^{ixu}\,du
\end{align}
where $F:=\frac{f\circ \phi^{-1}}{\phi'\circ\phi^{-1}}$ defined
on the interval $(\phi(a),\phi(b))$. Notice that
\begin{align}
\int^{\phi(b)}_{\phi(a)}|F(u)|\,du&=\int^{\phi(b)}_{\phi(a)}|f(\phi^{-1}(u))|\frac{1}{\phi'(\phi^{-1}(u))}\,du\\
&=\int^{\phi(b)}_{\phi(a)}|f(\phi^{-1}(u))|(\phi^{-1})'(u)\,du\\
&=\int^b_a|f(t)|\,dt<\infty
\end{align}
That is, $F\in L_1(\phi(a),\phi(b))$. Extending $F$ to $\mathbb{R}$ by setting $F=0$ outside the interval $(\phi(a),\phi(b))$ yields a function $F\in L_1(\mathbb{R})$. The conclusion the follows from the Riemann-Lebesgue theorem for $L_1(\mathbb{R})$ functions.
If there are a countable number of critical points for $\phi$, that is points where $\phi'=0$ within the interval; $(a, b)$, one can split the interval in subintervals $(a_n,a_{n+1})$ where $\phi\neq0$ ($\phi'(a_n)=0$ for all $n$) and apply the result to $\int^{a_{n+1}}_{a_n}f(t)e^{-ix\phi(t)}\,dt$. When the number of critical points is infinite, dominated convergence will justify that the validity of this splitting.