If $f$ is integrable, $\partial_x f$ makes sense in the distributional sense as $(\partial_x f,\phi)=(f,\partial_x \phi)$
Question. Does a sort of Taylor series make sense distributionally? that is, could the following be valid in any sense?
\begin{align} \left(\sum_{k=0}^{\infty}\frac{ (\partial_x)^k f}{k!},\phi\right)=\left(f,\sum_{k=0}^{\infty}\frac{(\partial_x)^k\phi}{k!}\right) \end{align}
Preliminary note : you forgot a minus sign in $(\partial_xf,\phi) = (f,-\partial_x\phi)$, coming from integration by parts.
Answer to the main question. Note that the operator $$ T_a := e^{a\partial_x} = \sum_{k=0}^\infty \frac{a^k}{k!} \frac{\partial^k}{\partial x^k} $$ is nothing else than the translation operator, such that $T_a(f)(x) = f(x+a)$. In consequence, the relation you propose, namely $$ \left\langle \sum_{k=0}^\infty \frac{1}{k!}\partial_x^kf, \phi\right\rangle = \left\langle f, \sum_{k=0}^\infty \frac{(-1)^k}{k!}\partial_x^k\phi \right\rangle, $$ is equivalent to $$ \int \overline{f(x+1)}\,\phi(x) \,\mathrm{d}x = \int \overline{f(x)}\,\phi(x-1) \,\mathrm{d}x, $$ which corresponds merely to the change of variable $x \mapsto x-1$.