Here's another stab.  It's based on the idea mentioned by Joel David Hamkins in his comments.

Let $a(x)$ and $b(x)$ be two such functions.  We'll use the fact that deciding whether or not $a(x)$ and $b(x)$ are identically equal is undecidable.  For this, we need a function $\Phi(a,b)$ that takes two functions and outputs $0$ if they are identically $0$ and some non-zero real otherwise.

One function we could take is simply $\Phi(a,b) = sup_{x} (a(x)-b(x))^2$.  Perhaps this is not allowable for your class of functions.  So I propose instead the function $$\Phi(a,b) = \int_{-\infty} ^{\infty} \frac{(a(x)-b(x))^2}{e^{|x|} (1+(a(x)-b(x))^2)} dx.$$

We then consider the integral of $e^{-t^2 \Phi}$, which is an elementary function iff $\Phi = 0$, which is iff $a(x)=b(x)$, which is undecidable.


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A slightly different punchline might be to consider the double integral

$$
\int \int \frac{(a(x)-b(x))^2}{e^{-|x|} (1+(a(x)-b(x))^2)} e^{-t^2} dx dt,
$$
which is perhaps not great because it's a function of two variables, which isn't likely what you had in mind.

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Or a third variation on this would be to define $\gamma(x) = |x|/x$ (and $\gamma(0) = 0$).  Then consider the function $\gamma((a(x)-b(x))^2)e^{-x^2}$, which has an elementary antiderivative iff the leading coefficient is $0$ almost everywhere (which is undecidable).

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A fourth variation is perhaps more satisfying.  We may assume $a(x)$ and $b(x)$ are in the ring generated by $\mathbb{Z}[x, \sin(x^n), \sin(x \sin(x^n))]$.  Let $C(x) = |a(x) - b(x)| - (a(x)-b(x))$.  Then [it is undecidable][1] to determine if $C(x)$ is identically $0$.  But for $C(x)$ of this form, it's all but certainly true that $e^{C(x) x^2}$ has an elementary antiderivative iff $C(x) = 0$.  [I am unsure how to prove this claim, but it could probably be proven along the same lines that $e^{ax^2}$ has no elementary antiderivative]


  [1]: https://en.wikipedia.org/wiki/Richardson's_theorem