Hamel basis is "not of this world":-)

It becomes more interesting if we make some mild a priori assumption,
like $f\in L^1(R)$, and $g_t^{(n)}\in L^1(R)$.
Then the answer is "yes", $f$ must be smooth.

We apply the Fourier transform. We obtain
$G_t(s)=(1-e^{its})F(s)$. As $g$ is $C^\infty$, we have that $|G_t(s)|\leq K_n|s|^{-n}$,
for every $n$.

Let us prove that $|F|=O(|s|^{-n})$ for an arbitrary fixed $n$.
Let $h(s)=s^nF(s)$. Suppose by contradiction that there is a sequence
$s_k\to\infty$ such that $h(s_k)\to\infty$.

As we have the inequality $|h(s)|\leq |C_t/\sin(st)|$
for ALL $t$, the sequence $s_k$ must have a strange property:
it must be close to ALL arithmetic progressions.

Does such a sequence exist?
Precise formulation: does these exist a sequence $s_k\to\infty$ such that for
every positive $a$ the sequence $S_k=\min_m|s_k-am|$ tends to $0$ ?

No it does not. Suppose it exists, and consider the integrals
$$I_k=\int_1^2 \min_m|s_k-mx|dx.$$
The function under the integral sign is bounded (does not exceed $1$). If our sequence
has the desired property, then $I_k\to 0$ (dominated convergence). But this is a
contradiction because the integrals evidently $\geq 1$.

The argument in the previous paragraph is due to user fedja.

It follows from this proof that positive measure set of $t$ is enough,
while any countable set
of $t$ is not enough.

EDIT. It is funny what happens if we require bounded support of $f$, in addition to $L^1$.
Then $F$ is an entire function of exponential type, bounded on the real line, so by Bernstein's theorem, $F'$ is bounded and we obtain that $f$ is smooth if $g_t$ is smooth for only ONE value of $t$.

EDIT.

Conjecture: if $f$ is locally $L^1$ and there is a set $E$ of positive measure such
that $g_t$ are smooth for $t\in E$ then $f$ is smooth. Same is probably true with $E$ of second Baire category.