Note that the function $\varphi(x) = \prod_{0\le k\le 2020}(x+k)$ is increasing on $[0,\infty)$ with $\varphi(0)=0$. Thus it has exactly one positive solution. I show the RHS, though I don't see the LHS argument yet. Suppose for the sake of the contradiction that $x_0>(2020!+6)^{-1}$. Now observe that on the one hand, $$ (x_0+1)\cdots(x_0+2020) =\frac{1}{x_0} <2020!+6. $$On the other hand $$ (x_0+1)\cdots(x_0+2020) =2020!+2020!\left(\sum_{1\le k\le 2020}\frac1k\right)x_0 +\cdots>2020!+2020!\left(\sum_{1\le k\le 2020}\frac1k\right)x_0. $$We now estimate $$ \sum_{1\le k \le 2020}\frac1k >1+\frac12+2\cdot \frac14 + 4\cdot\frac18+8\cdot\frac{1}{16}+\cdots+512\cdot \frac{1}{1024}+\frac{996}{2020}\triangleq 6+t. $$Now notice that this yields $$ (x_0+1)\cdots(x_0+2020)>2020!+\frac{2020!}{2020!+6}(6+t). $$Since $t$ is clearly bigger than $36/2020!$, this yields $(x_0+1)\cdots(x_0+2020)>6+2020!$, which is a contradiction.

We denote $H(k)=\sum_{i=1}^k \frac 1k$ and recall some classical inequalities: $$\frac{x}{1+x}<\ln{(1+x)}<x$$for any $x>0$ (see here: https://math.stackexchange.com/questions/652581/showing-fracx1x-log1xx-for-all-x0-using-the-mean-value-theorem) and it's consequence $$\ln{(k+1)}<H(k)<1+\ln k$$for any positive integer $k$. Let $k$ be equal to $2020$, $x_0=(k!+a)^{-1}$. Now we are ready to estimate $a$ frome above. Firsly, we rewrite original equality in the next way $$\prod_{i=1}^{k} \left(1+\frac{x_0}{i}\right ) =\frac{1}{k!x_0}=\frac{k!+a}{k!},$$$$\sum_{i=1}^k \ln {\left (1+\frac{x_0}{i} \right)} = \ln {\left ( 1+\frac{a}{k!} \right )}.$$We know that $$\sum_{i=1}^k \ln {\left (1+\frac{x_0}{i} \right)} <\sum_{i=1}^k \frac{x_0}{i}=x_0H(k)<x_0(1+\ln k)$$and $$\ln {\left ( 1+\frac{a}{k!} \right )}>\frac{a/k!}{1+a/k!}=\frac{a}{k!+a}=ax_0,$$so $x_0(1+\ln k)>ax_0$, or $a<1+\ln k\approx 8.611$.