Is the potential in Schrödinger equation an operator?

In the representation given by you, the $V(x,t)$ is a scalar function depending on $x$ and $t$. However, you already have choses a basis (the $x$-basis). In general, the $\hat{V}$ or rather $\hat{H}$ in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can write $\langle x |\hat{V}|x\rangle=V(x)$. Correspondingly, you choose the same basis for your state $|\psi\rangle$ with $\psi(x)=\langle x|\psi\rangle$.

The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction: $$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$ It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector operators like the position or momentum - but it is not a scalar in the Hilbert-space sense.

To answer the question I think you meant to ask, $V(x,t)$ is a real-valued function. But you can also think of it as an operator. An operator is anything that maps functions to other functions, and multiplication by a fixed function is one way to do that.

To be a little more pedantic with the notation, if you use $\hat{V}$ to represent the operator, then the action of this operator is defined as

$$\hat{V}f(x,t) = V(x,t)f(x,t)$$

for any $f(x,t)$ for which this equation makes sense.

The peculiar feature of these multiplicative operators is that they are diagonal in the position basis, so that $\langle x\rvert\hat{V}\lvert y\rangle \propto \delta(x - y)$. This means that, in a rough sense, a multiplicative operator like $\hat{V}$ doesn't do anything "weird" or "interesting" to the position-space function it's acting on - unlike a derivative operator, for example. (Of course things are different if you work in another basis.)