It is well known that the exercise of logic never adds to our knowledge: its role is to make a certain aspect of that knowledge clearer or more explicit, while keeping all the rest conveniently out of our sight.
This is a quote by Tommaso Toffoli, in "Entropy? Honest!". Entropy 18, 247 (2016). Reading your question reminded me of it because it is pretty much the point: we don't need to write our equations in tensorial form, we could indeed just write them in terms of each individual component (important note: those would not be scalars, for they wouldn't transform as scalars). However, this often will hide interesting properties of what we are dealing with that might make our life considerably easier (see, e.g., this brilliant answer by Terence Tao on a similar question in Math Overflow).
In the case of tensors, our main interest is that their components have well defined transformation properties under changes of coordinates and are, deep down, geometrically invariant, and that allows you to see and comprehend much more than you otherwise would.
For example, take the expression $\mathbf{F} = m \mathbf{a}$ in Classical Mechanics. This is written in terms of rank-1 tensors (vectors), but we could write it as three equations in components. It would read
$$
\left\lbrace
\begin{aligned}
F_x &= m a_x, \\
F_y &= m a_y, \\
F_z &= m a_z,
\end{aligned}
\right.
$$
where I employed Cartesian coordinates. Let us suppose you already measured $F_x$, $F_y$, and $F_z$ experimentally and now wants to determine the acceleration in each direction. However, you notice you messed up your setup and actually you wanted the components of the acceleration in a slightly different angle, rotated in $45º$ around the $z$ axis with respect to the system you chose. Then you compute the transformation and finds that the previous equations read now, after we transform them,
$$
\left\lbrace
\begin{aligned}
\frac{\sqrt{2}}{2} F_x - \frac{\sqrt{2}}{2} F_y &= m a_{x'}, \\
\frac{\sqrt{2}}{2} F_x + \frac{\sqrt{2}}{2} F_y &= m a_{y'}, \\
F_z &= m a_{z'},
\end{aligned}
\right.
$$
assuming I didn't get anything wrong. Had you chosen to work with vectors, the equation would read
$$\mathbf{F} = m \mathbf{a},$$
because while vector components transform under a change of basis, vectors themselves are geometrical objects which do not depend on the basis.
Of course, this is just an example, you'd have to compute the components to get to your final computation either way. However, notice how writing in vector notation employs automatically the invariance properties of vectors. You can choose to write component-wise, but that won't destroy the vectors that lie beneath your computations, it just hides them. The components still transform as vectors and the vectors are still there, you are just not looking at them, and, as a consequence, you are missing some information that could make things way simpler. The point is not that they are a way of organizing calculations (you'll often have to work with the components sooner or later), but that there are "hidden" symmetries and properties that become explicit once you cast the formulae into an appropriate notation.
At this point, it is worth pointing out the comment made by Dvij D.C. (which I'm adding here in case it is deleted in the future)
Tensors not only make the underlying symmetries more manifest but they are the only basis-independent/coordinate-independent way of expressing physics. The tensor-components $T_{\mu\nu}$ transform under a change of basis but the tensor itself $\textbf{T} = T_{\mu\nu} \textbf{e}^{\mu} \otimes \textbf{e}^{\nu}$ remains invariant.
Notice how this is similar to us writing Newton's second law in components and in vector form. Vectors are simply rank-1 tensors, and when employing them our equations automatically are expressed in an invariant way under rotations.
As a second example, consider the expression $\mathbf{p \cdot q}$ written in two other different notations. The first one, in Einstein notation, which would read
$$\mathbf{p \cdot q} = p_i q^i,$$
and the second being the expression in Cartesian coordinates,
$$\mathbf{p \cdot q} = p_x q_x + p_y q_y + p_z q_z.$$
Both expressions are tautological. Neither of them adds any intrinsic knowledge to us. However, the expression in Einstein notation makes it clear that the object is invariant under rotations, while the second one has a lot of objects that transform in a sort of complicated way, and you are not sure whether the expression would change if you chose to work with different coordinates.
In principle, one can component-wise everything we do with tensors. It is similar to searching for a needle in a haystack: you can do it by looking carefully and eventually you'll find it, but you can also use the extra knowledge that the needle is made of iron and use a magnet.
Extra Examples
Quantum Mechanics
One other example of the usage of tensors in Physics, although in a particularly specific notation, is within Quantum Mechanics. For simplicity, I'll stick to the rotation example. Suppose you made a measurement, noticed you should have rotated your coordinates before, and so on. If you denote the states in terms of wavefunctions, you'll figure out that the probability a particle prepared in the state $\psi(x)$ to be measured in the state $\phi(x)$ is
$$P(\phi|\psi) = \left\vert \int \phi^*(x) \psi(x) \mathrm{d}^3{x} \right\vert^2$$
before the rotation and, after the rotation,
$$P'(\phi|\psi) = \left\vert \int \phi'^*(x) \psi'(x) \mathrm{d}^3{x} \right\vert^2,$$
where $\psi'$ is the wavefunction $\psi$ after the rotation of the system of coordinates. Opening up the appropriate expression for $\psi'$ and $\phi'$, one will eventually find out that the probability is, of course, the same. In Dirac notation, however, which employs the fact that the objects we're dealing with are vectors, one would write
$$P(\phi|\psi) = \vert\langle \phi \vert \psi \rangle\vert^2$$
before the rotation and
$$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2.$$
Since rotations are implemented in Quantum Mechanics by means of unitary operators, one has then that $\vert\psi'\rangle = U \vert\psi\rangle$ for some unitary operator $U$. Hence, the rotated version is
$$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2 = \vert\langle \phi \vert U^\dagger U \vert \psi \rangle\vert^2 = \vert\langle \phi \vert \psi \rangle\vert^2 = P(\phi|\psi),$$
and the result is immediate using properties of linear algebra, without the need to manipulate anything with calculus.
Another example is how one can solve the quantum harmonic oscillator in terms of ladder operators instead of solving differential equations. It exploits the linear structure hidden in the wavefunctions and allows one to get a solution with much less labor.
Relativity
Expressions written in terms of tensors in Relativity are assured to work in all reference frames, since tensors are geometrical objects. This is similar to my previous example of how working in vector notation made taking the rotations into account much easier.
Take, for example, the famous expression
$$E^2 = p^2 c^2 + m^2 c^4.$$
In this form, it is not obvious that this formula holds in any inertial frame (unless, of course, you are already well acquainted with Relativity and know the fact by heart). However, the same formula can be written as
$$p^\mu p_\mu = m^2 c^2,$$
where $p^\mu = (\frac{E}{c}, \mathbf{p})^\intercal$, which explicitly shows the expression is invariant.
Another example is the fact that the relativistic Doppler effect for a source in uniform linear motion boils down to noticing that $k^\mu u_\mu$ is an invariant, $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$ being the wave's four-momentum and $u^\mu$ being the source's four-velocity. By computing the invariant in two different frames, one of which has the source at rest, one arrives at the expression for the frequency shift in an incredibly straightforward way. Indeed, suppose that in the frame of rest of the source the wave has $k^\mu = (\frac{\omega_0}{c}, \mathbf{k}_0)^\intercal$, while in some other frame the source has $u^\mu = (\gamma c, \gamma \mathbf{v})^\intercal $ and the wave has $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$. Then
$$
\begin{align}
k^\mu u_\mu\vert_{\text{rest}} &= k^\mu u_\mu\vert_{\mathbf{v}}, \\
- \omega_0 &= \gamma(-\omega + \mathbf{k \cdot v}).
\end{align}
$$
Let's write $\mathbf{k} = \frac{\omega}{c} \mathbf{\hat{n}}$, where $\mathbf{\hat{n}}$ is a unit vector (the fact that this is possible comes from usual wave mechanics). Then
$$
\omega_0 = \omega\gamma\left(1 - \frac{\mathbf{n \cdot v}}{c}\right),
$$
and we conclude that
$$\frac{\omega}{\omega_0} = \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{\mathbf{n \cdot v}}{c}}.$$
This derivation just comes in such a direct way because we noticed from the start that $k^\mu u_\mu$ is a relativistic invariant, something that is favored by tensor notation: the absence of "free" indices tells us that the quantity is an invariant.
