Yes, and under weaker conditions.

Set $s = (1-e^x) > 0$, and multiply your partial sums by $s$; this has no effect on log concavity. Then the resulting sequence is the Hadamard product of the coefficients of $(\sum_{i=1}^n a_i)$ with $(s^n)$, each of which is log concave (the first is the convolution of $(1,1,1,\dots,)$ with $(a_k)$), so the outcome is.

This argument requires only that $(a_k)$ be log concave (no monotonicity, etc). [There is a quick reduction to $(a_k)$ being a finite sequence if you don't like the possibly infinite convolution, since the $n$th partial sum only involve the first $n$ terms of $(a_k)$, and log concavity is checked just using the $n-1,n+1,n$th terms for each $n$.]

Yes, and under weaker conditions.

Set $s = (1-e^x) > 0$, and multiply your partial sums by $s$; this has no effect on log concavity. Then the resulting sequence is the Hadamard product of the coefficients of $(\sum_{i=1}^n a_i)$ with $(s^n)$, each of which is log concave (the first is the convolution of $(1,1,1,\dots,)$ with $(a_k)$), so the outcome is.

This argument requires only that $(a_k)$ be log concave (no monotonicity, etc). [There is a quick reduction to $(a_k)$ being a finite sequence if you don't like the possibly infinite convolution, since the $n$th partial sum only involve the first $n$ terms of $(a_k)$, and log concavity is checked just using the $n-1,n+1,n$th terms for each $n$.]

Edit: The original question has been edited (and not by the original proposer) to be quite different. The answer here is to what I thought was the original question, which was whether $(\sum_{i=1}^n a_i (1-e^x)^{n-1})$ is a log concave sequence in $n$.