This is a follow-up to a much simpler question I asked here, which @PrincessEev answered promptly and perfectly. She showed me how to rewrite the sum $\sum _{i=1}^x \phi (x-i)$ in such a way that Abel Summation could be applied, to obtain
$$\sum _{i=1}^x \phi (x-i) = \lfloor x \rfloor \phi(x-1) – \int_0^{x-1} \lfloor u+1 \rfloor \phi'(u) \, \mathrm{d}u$$
Here is a harder problem – I think. And once again, I don't really know where to start. Let $1 \le j \le x$ be a constant positive integer. Can one apply Abel Summation to the sum $\sum _{i=1}^x \phi (x – i j)$? If so, how?
CLARIFICATION / UPDATE
I initially added follow-up material here asking what I was doing wrong in implementing @junjios's answer. The fault was mine and was a coding error in my use of Mathematica.
Since it has no bearing on @junjios's correct answer, I have deleted those elements of this post, to avoid confusing new readers. If you are puzzled over some of the posts in the comments section, this is the explanation.
Best Answer
Let $x\geq 1$ and $j\in\mathbb{R}$ be fixed. (Note: There are no further conditions on $j$!)
We define: $\psi(y):=\phi(x-yj)$. Clearly, $\psi$ is continuously differentiable, so we can apply the Abel summation formula to obtain $$\sum_{i=1}^x\phi(x-ij)=\sum_{i=1}^x\psi(i)=\lfloor x\rfloor\psi(x)-\int_1^x\lfloor u\rfloor\psi'(u)\,\mathrm{d}u\\=\lfloor x\rfloor\phi(x(1-j))+j\int_1^x\lfloor u\rfloor\phi'(x-uj)\,\mathrm{d}u.$$
Generalisation:
More generally, given any (further) continuously differentiable function $f$ (possibly depending on $x$), we may write $$\sum_{i=1}^x\phi(f(i))=\lfloor x\rfloor\phi(f(x))-\int_1^x\lfloor u\rfloor\phi'(f(u))f'(u)\,\mathrm{d}u.$$