Here's a sketch of a proof of a generalization: $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}.\tag {$*$}$$ (This is a generalization of Terry Tao's generalization, which is the case $a=0$.)

We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways.

Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.

Here's a sketch of a proof of a generalization:

\begin{multline} \quad \sum_{k=1}^n\binom nk \frac{t^k}{k+a}\\ =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}. \quad \tag {$*$} \end{multline}

(This is a generalization of Terry Tao's generalization, which is the case $a=0$.)

We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways.

Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.

Here's a sketch of a proof of a generalization: $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}.\tag {$*$}$$ (This is a generalization of Terry Tao's generalization, which is the case $a=0$.)

We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways.

Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each of the sums has no constant term in $t$, and $(*)$ follows.

Here's a sketch of a proof of a generalization: $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}.\tag {$*$}$$ (This is a generalization of Terry Tao's generalization, which is the case $a=0$.)

We start with the identity $$\sum_{k=0}^n \binom nk \frac{t^k}{k+a} = \frac {1}{a\binom{a+n}{n}}\sum_{k=0}^n \binom{a+k-1}{k} (1+t)^k.$$ This is a special case of a well-known linear transformation for the hypergeometric series, the case $b=a+1$ of $${}_2F_1(-n,a; b\mid -t) =\frac{(b-a)_n}{(b)_n}\,_2F_1(-n,a; 1-n-b+a\mid 1+t),$$ where $(u)_n = u(u+1)\cdots (u+n-1)$, which can be proved easily in several ways.

Since $\frac{1}{a}\binom{a+k-1}{k} = \frac {1}{k}\binom{a+k-1}{k-1}$ for $k\ge 1$, we have $$\sum_{k=1}^n\binom nk \frac{t^k}{k+a} =\frac{1}{\binom{a+n}{n}}\sum_{k=1}^n \binom {a+k-1}{k-1} \frac{(1+t)^k-1}{k}+C$$ where $C$ is a constant (as a polynomial in $t$). But $C=0$ since each summand has no constant term in $t$, and $(*)$ follows.