Solution:

Let the number of cards be $c$ and let the number of distinct names be $n$. Each card contains a set of names; denote these sets by $S_{1}, S_{2}, \ldots, S_{c}$ (some of these sets may share elements). Now let $E$ be a subset of the set of $n$ names, and denote by $D(E)$ the difference of the number of cards in the left stack and the number of cards in the right stack, after the cards are shuffled by the names in $E$. We are given that $D(\emptyset)>0$; we must show that there exists a subset $E$ such that $D(E)<0$.

Consider the sum of all $D(E)$ as $E$ ranges through all $2^{n}$ possible subsets of names. If we can show that this sum is equal to zero, we will be done, since $D(\emptyset)>0$ is just one term in this sum, forcing at least one other term to be negative. For $i=1,2, \ldots, c$, let $v_{i}=+1$ or $-1$ if the $i$th card is initially in the left or right stack, respectively. Then
$$D(\emptyset)=v_{1}+v_{2}+\cdots+v_{c}=\sum_{i=1}^{c} v_{i}$$
Now consider what happens when we shuffle the cards corresponding to a subset $E$ of names. The $i$th card will move back and forth from one stack to the other a total of $|E \cap S_{i}|$ times ($|A|$ means the number of elements in the set $A$). The only thing that matters is whether this value is even or odd. Thus we have
$$D(E)=\sum_{i=1}^{c}(-1)^{|E \cap S_{i}|} v_{i}$$
It remains to sum this expression over all subsets $E$. Let us examine what happens just for one card; i.e., let us compute
$$\sum_{E}(-1)^{|E \cap S_{i}|} v_{i}$$
for a fixed $i$ as $E$ ranges over all $2^{n}$ subsets of names. If we can show that this equals zero, then the entire sum will equal zero and we are done. Let $|S_{i}|=k$. Then $|E \cap S_{i}|$ will range from $0$ to $k$ inclusive. For $0<r<k$, how many subsets $E$ are there such that $|E \cap S_{i}|=r$? There are $\binom{k}{r}$ subsets of $S_{i}$ with $r$ elements. Fix one of them, call it $T$. Then $E$ must contain all the elements of $T$, plus any subset of the names that are not contained in $S_{i}$. In other words, there are $2^{n-k}$ subsets containing $T$, and thus there are $\binom{k}{r} 2^{n-k}$ subsets $E$ altogether satisfying $|E \cap S_{i}|=r$. Therefore
$$\sum_{E}(-1)^{|E \cap S_{i}|} v_{i}=v_{i} \sum_{r=0}^{k}(-1)^{r}\binom{k}{r} 2^{n-k}=2^{n-k} v_{i} \sum_{r=0}^{k}(-1)^{r}\binom{k}{r}$$
By the binomial theorem, we have
$$2^{n-k} v_{i} \sum_{r=0}^{k}(-1)^{r}\binom{k}{r}=2^{n-k} v_{i}(1-1)^{k}=0$$

We will prove the statement by induction on $n$, the number of distinct names present. Call the left stack $L$ and the right stack $R$. For the case $n=1$, let $L$ have $x$ cards and $R$ have $y$ cards with $x>y$; all cards have the same single name written on them. We perform the shuffle corresponding to the single name present, and we are done (now we have $x$ in $R$ and $y$ in $L$).

Next, assume that we have proven the statement for $n$ names, and consider the case with $n+1$ names. Call the first $n$ names $a_{1}, a_{2}, \ldots, a_{n}$, and let the new $(n+1)$-st name be $a$. There are two cases:

Case 1. The number of $L$ cards containing only name $a$ is less than or equal to the number of $R$ cards containing only name $a$. In this case, we need only look at the remaining cards (the ones that do not contain only $a$). We can ignore the name $a$, and use the inductive hypothesis to perform the required shuffles using some subset of the $n$ names $a_{1}, a_{2}, \ldots, a_{n}$, and we will be done: there are now more of these remaining cards in stack $R$ than in stack $L$, and since there were at least as many "only $a$" cards in $R$ than in $L$, the final configuration has more cards in $R$ than in $L$. (Note that if the original configuration contains as many "remaining" cards in $R$ as in $L$, we don't have to perform any shuffles at all, given that the "only $a$" cards in $R$ are more than those in $L$.)

Case 2: The number of $L$ cards containing only name $a$ is greater than the number of $R$ cards containing only $a$. In this case, we first perform the shuffle corresponding to the name $a$. This will switch the "only $a$" cards, so that there are more of them in $R$ than in $L$. Unfortunately, some of the remaining cards may also have moved around. But now there are two possibilities: if there are at least as many remaining cards in $R$ than in $L$, we are done. If not, then we can again employ the inductive hypothesis, ignoring the name $a$, and just applying the shuffles to some subset of the $n$ names $a_{1}, a_{2}, \ldots, a_{n}$. The end result will be that there will be as many remaining cards in $R$ than in $L$, and we are done.