The problem is that, with an elasticity of substitution $\sigma<1$, in the CES production function we have negative exponents.$^1$
I rewrite your CES production function:
$$Y = f(K, L) = (a \cdot K^\rho + (1 - a) \cdot L^\rho )^{1/\rho}.\;\;\;\;\;(1)$$
Elasticity of substitution $\sigma = 1/(1-\rho) <1$ implies $\rho<0$, so that the exponents $\rho$ and $1/\rho$ in equation $(1)$ are negative.
In particular, this implies that the factors $K$ and $L$ in $(1)$ are the denominator of a fraction. Therefore, we cannot set $K=0$ or $L=0$.
To see it clearly, let’s rewrite $(1)$ as
$$Y = f(K, L) = (a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}} )^{1/\rho}\;\;\;(2)$$
where, obviously, $-\rho >0$ if $\rho <0$.
Even though we cannot set $K$ or $L$ equal to $0$ to show that the factors are all essential, we can, instead, take the limit of the production function as $K$ or $L$ go to 0.
For example, let's take the limit of $(1)$ as $L \rightarrow0$ (for a fixed level of $K$).
Rewrite $(2)$ as
$$Y = f(K, L) = \frac {1}{(a \cdot \frac {1} {K^{-\rho}} + (1 - a) \cdot \frac {1} {L^{-\rho}} )^{-1/\rho}}\;\;\;\;(3)$$
(remember that $-\rho$ and $-1/\rho$ are positive numbers).
As $L\rightarrow0$ then $1/L^{-\rho}\rightarrow \infty$, and the overall denominator of $(3)$ goes to $\infty$. Therefore, the fraction $(3)$ representing the production $Y$ goes to $0$.
That is, as $L \rightarrow 0$ production $Y \rightarrow 0$, irrespective of the kevel of the other factor $K$.
A similar, symmetric, argument applies to $K$.
Hence, both factors are essential, because any of them going to zero implies that production goes to zero.
$^1$ And we can clash with the awful mathematical prohibition: “You cannot divide by zero”, the mortal sin in mathematics 😊.
