The idea is to dominate $e^{sx}$ with a quadratic function. To achieve this, we exploit the property that all higher derivative of $e^{sx}$ are positive. [A search shows that this property is called completely monotonic.]
Lemma: Let $\Phi(x) = e^{sx}$ and $\Psi(x) = a_2 x^2 + a_1 x + a_0$ be a quadratic function. If we fit $\Psi$ for some $a < b$ such that $\Phi(a) = \Psi(a)$, $\Phi(b) = \Psi(b)$, and $\Phi'(a) = \Psi'(a)$, then $\Phi(x) \leq \Psi(x)$ on $[-\infty, b]$.
Proof: We consider the Taylor expansion of $\Phi$ and $\Psi$ at $a$. We have
$$\Phi(x) = \Phi(a) + \Phi'(a)\cdot (x - a) + \frac{1}{2}\Phi''(a)\cdot (x - a)^2 + \cdots$$
while since $\Psi$ is quadratic, we have
$$\Psi(x) = \Psi(a) + \Psi'(a)\cdot (x - a) + \frac{1}{2}\Psi''(a)\cdot (x - a)^2.$$
In particular, taking $x = b$ gives
$$\frac{1}{2}\Psi''(a)\cdot (b - a)^2 = \frac{1}{2}\Phi''(a)\cdot (b - a)^2 + \sum_{n \geq 3} \frac{1}{n!}\Phi^{(n)}(a)\cdot (b - a)^n$$
as the higher derivative are positive, we have $\Psi''(a) > \Phi''(a)$.
This immediately implies the inequality for $x < a$, since in this case
$$\Phi(x) = \Phi(a) + \Phi'(a)\cdot (x - a) + \frac{1}{2}\Phi''(a)\cdot (x - a)^2 + \frac{1}{6}\Phi'''(\xi)\cdot (x - a)^3$$
and the third term is negative, so
$$\Phi(x) \leq \Phi(a) + \Phi'(a)\cdot (x - a) + \frac{1}{2}\Phi''(a)\cdot (x - a)^2$$
which is at most
$$\Psi(x) = \Psi(a) + \Psi'(a)\cdot (x - a) + \frac{1}{2}\Psi''(a)\cdot (x - a)^2.$$
For $x \in [a, b]$, we have
$$\frac{1}{2}\Psi''(a)\cdot (b - a)^2 = \frac{1}{2}\Phi''(a)\cdot (b - a)^2 + \sum_{n \geq 3} \frac{1}{n!}\Phi^{(n)}(a)\cdot (b - a)^n$$
thus
$$\frac{1}{2}\Psi''(a)= \frac{1}{2}\Phi''(a) + \sum_{n \geq 3} \frac{1}{n!}\Phi^{(n)}(a)\cdot (b - a)^{n - 2}.$$
We now note that replacing $b$ with $x$ decreases RHS since every term is positive, so
$$\frac{1}{2}\Psi''(a) \geq \frac{1}{2}\Phi''(a) + \sum_{n \geq 3} \frac{1}{n!}\Phi^{(n)}(a)\cdot (x - a)^{n - 2}$$
which translates back to $\Psi(x) \geq \Phi(x)$.
Thus we have established the lemma. The rest is straightforward. Write $u = \sigma^2 / c^2$. Take $a = -cu$ and $b = c$. The quadratic $\Psi$ exists since it is determined by three linear equations, thus
$$\mathbb{E} \Phi(X) \leq \mathbb{E} \Psi(X) = a_2 \sigma^2 + a_0.$$
We have
$$a_2 c^2 u^2 - a_1 cu + a_0 = \Phi(-cu),$$
$$a_2 c^2 + a_1 c + a_0 = \Phi(c),$$
so taking a linear combination
$$a_2 \sigma^2 + a_0 = a_2 c^2u + a_0 = \frac{u}{u + 1} \Phi(c) + \frac{1}{u + 1} \Phi(-cu).$$
Thus we conclude the desired result
$$\mathbb{E} \Phi(X) \leq \frac{u}{u + 1} \Phi(c) + \frac{1}{u + 1} \Phi(-cu)$$
and the equality is achieved when $X$ is supported on $\{c, -cu\}$.
