Here are the calculations steps, in case it can help anyone:

Left hand side can be written
$$e^{t(a+b)} = \sum_{k} \frac{(a+b)^k t^k}{k!}$$
where one recognizes the term we want to compute.

Right hand side is given by
$$e^{ta} e^{tb} e^{- c t^2 / 2} = \sum_{i, j, k} t^{i + j + 2k} \frac{a^i b^j (-c / 2)^k}{i! \, j! \, k!}$$

Identifying both sides, one has:
$$\frac{(a+b)^n}{n!} = \sum_{i + j + 2k = n} \frac{(-c / 2)^k}{i! \, j! \, k!} a^i b^j$$

$$\frac{(a+b)^n}{n!} = \sum_{\substack{i + j \leq n \\ i+j = n [2]}} \frac{(-c / 2)^{(n - i - j) / 2}}{i! \, j! \, \left(\frac{n - i - j}{2}\right)!} a^i b^j$$

Let us note $k = m+n$ and $r = m$,
$$\frac{(a+b)^n}{n!} = \sum_{\substack{0 \leq r \leq k \leq n \\ k = n [2]}} \frac{(-c / 2)^{(n - k) / 2}}{\left(\frac{n - k}{2}\right)! \, r! \, (k-r)!} a^r b^{k-r}$$

$$\frac{(a+b)^n}{n!} = \sum_{\substack{k = 0 \\ k = n [2]}}^n \frac{(-c / 2)^{(n - k) / 2}}{k! \left(\frac{n - k}{2}\right)!} \sum_{r = 0}^{k} \binom{k}{r} a^r b^{k-r}$$

Which is the expected formula.