I think I can write down an expression for the probability, but I'm not sure how useful it is. My interpretation of the random experiment is that a random $(S\times L)$-matrix over $\{1,\ldots,N\}$ is generated by choosing the entries independently and uniform, and the $i$-th multiset is then simply the $i$-th row of this matrix. Clearly, the probability for every single matrix is $1/N^{S\cdot L}$. A matrix is good if every row contains at least $k$ entries that do not appear in any other row, so it remains to count the good matrices. Any good matrix induces a partition $\{1,\ldots,N\}=B_0\cup B_1\cup\cdots\cup B_S$, where $B_i$ for $i=1,\ldots,S$ is the set of elements occuring only in row $i$ (in particular nonempty). The number of good matrices corresponding to a given partition is $$\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}|B_i|^K|B_0|^{L-K}.$$ Now let $\mathcal P$ be the set of ordered partitions $N=b_0+b_1+\cdots+b_S$ of $N$ into nonnegative integers, where $b_i\geqslant 1$ for $i=1,\ldots,S$. Then the total number of good matrices can be written as $$A=\sum_{(b_0,b_1,\ldots,b_S)\in\mathcal P}\binom{N}{b_1}\binom{N-b_1}{b_2}\cdots\binom{N-b_1-\cdots-b_{S-1}}{b_S}\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}b_i^Kb_0^{L-K},$$ and your probability is $A/N^{S\cdot L}$.

Edit: Changed the formulas slightly, making it (hopefully) correct, but even less useful.

I think I can write down an expression for the probability, but I'm not sure how useful it is. My interpretation of the random experiment is that a random $(S\times L)$-matrix over $\{1,\ldots,N\}$ is generated by choosing the entries independently and uniform, and the $i$-th multiset is then simply the $i$-th row of this matrix. Clearly, the probability for every single matrix is $1/N^{S\cdot L}$. A matrix is good if every row contains at least $k$ entries that do not appear in any other row, so it remains to count the good matrices. Any good matrix induces a partition $\{1,\ldots,N\}=B_0\cup B_1\cup\cdots\cup B_S$, where $B_i$ for $i=1,\ldots,S$ is the set of elements occuring only in row $i$ (in particular nonempty). The number of good matrices corresponding to a given partition is $$\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}S(K,|B_i|)\cdot |B_i|!\cdot |B_0|^{L-K},$$ where $S(K,|B_i|)$ is the Stirling number of the second kind, the number of partitions of a $K$-set into exactly $|B_i|$ nonempty subsets.

Now let $\mathcal P$ be the set of ordered partitions $N=b_0+b_1+\cdots+b_S$ of $N$ into nonnegative integers, where $1\leqslant b_i\leqslant L$ for $i=1,\ldots,S$. Then the total number of good matrices can be written as $$A=\sum_{(b_0,b_1,\ldots,b_S)\in\mathcal P}\binom{N}{b_1}\binom{N-b_1}{b_2}\cdots\binom{N-b_1-\cdots-b_{S-1}}{b_S}\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}S(K,b_i)\cdot b_i!\cdot b_0^{L-K},$$ and your probability is $A/N^{S\cdot L}$.

I think I can write down an expression for the probability, but I'm not sure how useful it is. My interpretation of the random experiment is that a random $(S\times L)$-matrix over $\{1,\ldots,N\}$ is generated by choosing the entries independently and uniform, adn the $i$-th multiset is then simply the $i$-th row of this matrix. Clearly, the probability for every single matrix is $1/N^{S\cdot L}$. A matrix is good if every row contains at least $k$ entries that do not appear in any other row, so it remains to count the good matrices. Any good matrix induces a partition $\{1,\ldots,N\}=B_0\cup B_1\cup\cdots\cup B_S$, where $B_i$ for $i=1,\ldots,S$ is the set of elements occuring only in row $i$ (in particular nonempty). Now the number of good matrices corresponding to a given partition is $$\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}|B_i|^K|B_0|^{L-K}.$$ Now let $\mathcal P$ be the set of ordered partitions $N=b_0+b_1+\cdots+b_S$ of $N$ into nonnegative integers, where $b_i\geqslant 1$ for $i=1,\ldots,S$. Then the total number of good matrices can be written as $$A=\sum_{(b_0,b_1,\ldots,b_S)\in\mathcal P}\binom{N}{b_1}\binom{N-b_1}{b_2}\cdots\binom{N-b_1-\cdots-b_{S-1}}{b_S}\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}b_i^Kb_0^{L-K},$$ and your probability is $A/N^{S\cdot L}$.

I think I can write down an expression for the probability, but I'm not sure how useful it is. My interpretation of the random experiment is that a random $(S\times L)$-matrix over $\{1,\ldots,N\}$ is generated by choosing the entries independently and uniform, and the $i$-th multiset is then simply the $i$-th row of this matrix. Clearly, the probability for every single matrix is $1/N^{S\cdot L}$. A matrix is good if every row contains at least $k$ entries that do not appear in any other row, so it remains to count the good matrices. Any good matrix induces a partition $\{1,\ldots,N\}=B_0\cup B_1\cup\cdots\cup B_S$, where $B_i$ for $i=1,\ldots,S$ is the set of elements occuring only in row $i$ (in particular nonempty). The number of good matrices corresponding to a given partition is $$\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}|B_i|^K|B_0|^{L-K}.$$ Now let $\mathcal P$ be the set of ordered partitions $N=b_0+b_1+\cdots+b_S$ of $N$ into nonnegative integers, where $b_i\geqslant 1$ for $i=1,\ldots,S$. Then the total number of good matrices can be written as $$A=\sum_{(b_0,b_1,\ldots,b_S)\in\mathcal P}\binom{N}{b_1}\binom{N-b_1}{b_2}\cdots\binom{N-b_1-\cdots-b_{S-1}}{b_S}\prod_{i=1}^S\sum_{K=k}^L\binom{L}{K}b_i^Kb_0^{L-K},$$ and your probability is $A/N^{S\cdot L}$.