Main claim. $$\Pr[|X - \mathbb{E} X| \geq t] \leq \frac{\mathbb{E} X}{t^2} \approx \frac{L}{t^2}. $$ You can bound the variance and use Chebyshev's Inequality as you suggest, and the calculations are not pretty but it helps if someone has already done them. Here's a sketch. Let's write $m = |M|$ for the number of samples, and $N$ for the size of the set. Let $I_{i,j}$ equal $1$ if $M[i]=M[j]$ and zero otherwise, then $X = \sum_{i < j} I_{i,j}$, the number of collisions.
Let $A_x$ be the probability of drawing element $x$ from the set. I understand you are interested in the uniform case $A_x = \frac{1}{N}$ but I only know how to solve the general case.
Claim 1 (as you wrote). $\mathbb{E} X = {m \choose 2} \|A\|_2^2$. Proof: the probability that $M[i] = M[j]$ is $\sum_x A_x^2 = \|A\|_2^2$, then use linearity of expectation. For the uniform distribution, $\|A\|_2^2 = \frac{1}{N}$.
Claim 2. (I will sketch below.) $$ Var(X) = {m \choose 2}\left(\|A\|_2^2 - \|A\|_2^4\right) + 6 {m \choose 3} \left(\|A\|_3^3 - \|A\|_2^4\right) .$$ For the uniform distribution, $\|A\|_3^3 = \frac{1}{N^2}$, so the second term cancels and we get: $$ Var(X) = {m \choose 2}\left(\frac{1}{N} - \frac{1}{N^2}\right) \leq \mathbb{E} X . $$
Corollary 3. Now we can use Chebyshev's inequality, i.e. $$ \Pr[ |X - \mathbb{E} X| \geq t] \leq \frac{Var(X)}{t^2} \leq \frac{\mathbb{E} X}{t^2} . $$
The only thing for me to convince you of is Claim 2, which comes from a counting argument. Remember $I_{i,j}$ is the indicator for a collision between $i$ and $j$th samples.
\begin{align} Var(X) &= Var\left(\sum_{i<j} I_{i,j}\right) \\ &= \mathbb{E} \left(\sum_{i<j} (I_{i,j} - \mathbb{E}I_{i,j})\right)^2 \\ &= \sum_{i<j} \sum_{k<\ell} \mathbb{E} \left(I_{i,j} - \mathbb{E}I_{i,j}\right)\left(I_{k,\ell} - \mathbb{E}I_{k,\ell}\right) \\ &= \sum_{i<j} \sum_{k<\ell} \mathbb{E} I_{i,j} I_{k,\ell} - \mathbb{E} I_{i,j} \mathbb{E} I_{k,\ell} . \end{align}\begin{align} Var(X) &= Var\left(\sum_{i<j} I_{i,j}\right) \\ &= \mathbb{E} \left(\sum_{i<j} (I_{i,j} - \mathbb{E}I_{i,j})\right)^2 \\ &= \sum_{i<j} \sum_{k<\ell} \mathbb{E} \left(I_{i,j} - \mathbb{E}I_{i,j}\right)\left(I_{k,\ell} - \mathbb{E}I_{k,\ell}\right) \\ &= \left(\sum_{i<j} \sum_{k<\ell} \mathbb{E} I_{i,j} I_{k,\ell} \right) - 2 \left(\sum_{i<j} \sum_{k<\ell} \mathbb{E} I_{i,j} \mathbb{E} I_{k,\ell} \right) + \left(\sum_{i<j} \sum_{k<\ell} \mathbb{E} I_{i,j} \mathbb{E} I_{k,\ell} \right) \\ &= \sum_{i<j} \sum_{k<\ell} \mathbb{E} I_{i,j} I_{k,\ell} - \mathbb{E} I_{i,j} \mathbb{E} I_{k,\ell} . \end{align} (Note $\mathbb{E} I_{i,j} \mathbb{E} I_{k,l} = \|A\|_2^4$.) Now for each $i,j,k,\ell$ in this sum, we have three cases:
If $i=k, j=\ell$ then $I_{i,j} = I_{k,\ell}$ and $\mathbb{E} I_{i,j}I_{k,\ell} = \mathbb{E}I_{i,j} = \|A\|_2^2$. There are ${m \choose 2}$ such terms.
If $\left| \{i,j\} \cap \{k,\ell\} \right| = 1$, then we can calculate $\mathbb{E} I_{i,j}I_{k,\ell}$ is the probability that three independent samples are all the same element, which is $\sum_x A_x^3 = \|A\|_3^3$. We can count to get $6 {m\choose 3}$ such terms, because there are ${m\choose 3}$ triples of distinct indices, and each triple $a<b<c$ appears in the sum $6$ ways, namely $3$ ways to assign $i<j$ to two of $a,b,c$, times $3$ ways to assign $k<\ell$, minus the three combinations where $i=j$ and $k=\ell$.
If $\{i,j\} \cap \{k,\ell\} = \emptyset$, then the random variables $I_{i,j}$ and $I_{k,\ell}$ are independent and the term is zero. By the way, there are $6{m\choose 4}$ such terms because each choice of $4$ distinct indices $a<b<c<d$ appears six times (count three each with $a=i$ and with $a=k$). Now you can check I've counted all the terms in the sum, since there are ${m\choose 2} {m \choose 2}$ terms and this equals ${m\choose 2}$ (from case 1) plus $6{m\choose 3}$ (from case 2) plus $6{m\choose 4}$ (from case 3).
If this is not complete enough, I can link a reference but I prefer not to self-cite and do not know where else to find this proof (though as you said, it's unlikely to be unique).