Clearly, $P(0)=0$ is possible, since then $x=0$ is a root for any $a$. We now show that no other values work. Write $n=2022$ and $P(x)=\sum_{j=0}^{n-1} c_j x^j$. Since $P$ works iff $-P$ works, we assume wlog that $c_0 = P(0) > 0$. The idea is that all terms except $x^n$ and the constant term are essentially negligible for a suitable choice of $a$, and the sum of these terms is always positive. Indeed, take some small $a \in (0 , 1 )$. By AM-GM, we have for all $1 \le j \le n-1$ that $$ x^n + a = j \cdot \frac{x^n}{j} + (n-j) \cdot \frac{a}{n-j}\ge n \sqrt[n]{ |x^n / j | ^j (a / (n-j)) ^{n-j}} = \frac{n}{\sqrt[n]{j^j (n-j)^{n-j} } } |x|^j a^{ \frac{n-j}{n} } \ge |x|^j a^{ \frac{n-j}{n} } .$$Therefore, $$| a c_j x^j | \le | c_j | ( x^{n} a^{ j / n } + a ^{1 + j/n} ) = | c_j | a ^{ j / n } ( x^n + a )< \frac{x^n}{n} + \frac{a c_0}{n}, $$if we choose $a$ sufficiently small. This means that $$x^n + aP(x) = x^n + \sum_{j=1}^{n-1} a c_j x^j + a c_0 \ge x^n + ac_0 - \sum_{j=1}^{n-1} |a c_j x^j | > (x^n + a c_0) \left( 1 - \frac{n-1}{n} \right) > 0,$$for all $x$, hence no real root.