Several concepts are tangled up in your question.
The electrons in their orbitals are a stable solution to a potential problem, the potential supplied by the charges in the problem. The orbitals describe the probability of finding the electron in a position (x,y,z) around the nucleus.
How can one see the electron orbital? In this link you can see that it is not a simple matter, but it has been done, by disturbing the electron with photons and getting statistically the original location.
Once one has seen/measured the electron's location the solution of the potential no longer holds for this free electron. The probability of finding it can be described by a traveling plane wave, until it is caught by another potential and radiates down to a ground state or an energy level that was empty in an ionized atom.
Actually atoms get incident by electromagnetic radiation at every instant, if we assume electrons to get superposed by electromagnetic wave, electron can't exist as stationary wave, but according to schrodinger model, electron is said to exist as stationary wave.
A photon hitting a bound ( probability standing wave) electron can expel it from the atom . A photon of appropriate energy can kick it up to a higher energy level , the photon disappears and the electron falls back to ground state by emitting that special spectral line. It is true that atoms, and this means the electron cloud about the nucleus, are continuously hit and interacting with photons, but the grand majority of photons do not have enough energy even to ionise the atoms, let alone free the electron.
The photons in our environment are bounded in energy to first order by the black body radiation of the earth and by the spectrum from the sun which supplies the energy to the surface of the earth. The high tail from the sun which can ionize atoms and thus is dangerous is mostly cut off by the atmosphere, our shelters and clothes.
So the only two outcomes on atoms from the "superposition" you imagine are either change of energy level and disappearance of photon, or ionization of atom and change in energy of photon. This last does destroy the standing wave probability function of the electron to a plane wave one.
A free electron hit by a photon can either scatter elastically or inelastically, but a free electron is no longer described by a probability standing wave, just by a plane wave propagating ( again , a probability wave), which is also a solution of Schrodinger's equation in the absence of a potential.
"The wave is actually probability in the sense that it assigns probability to the space coordinates of detecting photon at a certain time."
No, the emergence of the classical EM wave from the quantum wavefunction of the photon is not trivial, because a classical EM wave is made up of many photons. In particular, it is not the case that the classical EM wave is the wavefunction of a photon. (Even more particular, it is difficult to even speak of the wavefunction of a photon, since photons usually arise in a quantum field theoretic ("second-quantized") description where the notion of wavefunction does not exist (but is, if you insist on something comparable, replaced by a wavefunctional))
Also, do not speak of "the electron wave". While an electron - like all quantum objects - carries wave-like properties and can be described by a wavefunction (which is not a function as you might imagine it if we incorporate its spin in the sense that it does not take values in the real or complex numbers), it isn't a wave in any classical sense, and also still carries particle-like properties. It is a quantum object, neither fully wave nor fully particle at any time.
Nevertheless, all quantum objects (and hence all things you might decsribe as "matter waves") of course carry energy - the energy that is in their rest mass and the energy that is in their momentum, though the energy of any given quantum state may not be well-defined, but "smeared out" over a range of energies.
To ask how physical objects carry the properties they do is, deep down, not sensible. How does a classical particle carry momentum? By having mass and velocity! But how does it carry mass and velocity? By...um...moving and stuff. How does it move? Um... You get the idea. "Why/How" is a question that can be asked infinitely many times, but only answered finitely many before you hit a point where the only answer is because it seems that way.
Best Answer
One must distinguish the underlying quantum mechanical framework from the emergent classical mechanics and electrodynamics framework when discussing waves.
In classical mechanics wave equations are solutions of differential equations which depend on the $(x,y,z,t)$ variables of massive ideal particles which are derivable from differential equations . These equations are called wave equations from the everyday terminology which first arose from water waves, and water waves can be described by a differential equation, called a wave equation. The solutions are sinusoidal functions which fit the waves in $(x,y,z,t)$ . Energy is transported by these waves, but there is not much mass motion. Classical electromagnetic waves also transport energy.
The confusion comes from the de Broglie interpretation which is applicable to the quantum mechanical level of elementary particles, atoms , molecules ... The de Broglie wavelength:
$$\lambda = \frac{h}{p}$$
This labeling as "matter waves" is what causes the confusion and should not really be used. The diffraction experiments with electrons show an associated wave interference pattern with the deposition of electrons on screens ( and measuring instruments) , and the de Broglie wavelength which depends on the momentum is verified by experiment. BUT the wave differential equation which describes accurately the behavior of the electrons is a second order differential equations which only predicts probabilities, according to the postulates of quantum mechanics.
So it is not that the electron is spread out all over the available phase-space space. It is the probability of the electron to be found at $(x,y,z,t)$ that controls the diffraction patterns of the experiment.
One always measures a whole electron, in any experiment, not fractions of it. Each dot is one whole electron in this single electron double slit experiment. It is the probability of finding it that is "waving" and generating the interference pattern. The end result looks like a matter wave, but it is a bad terminology, because no matter is waving.
Edit incorporating part of comments:
From matter transport the question in comments became of whether a quantum mechanical particle has a specific reality, trajectory.
Obviously classical trajectories exist otherwise bullets would not be sure to find the target. Mathematical functions of classical mechanics describe classical trajectories with great accuracy, macroscopically , i.e at dimension where the Heisenberg uncertainty principle is automatically fulfilled because of the large numbers involved. The HUP is in a way a mnemonic of quantum mechanical behavior, because it is the direct result of the existence of non zero commutator relations which are at the basis of quantum mechanical theory. The HUP gives an envelope in the two non commuting variables, for example momentum and position, which are relevant for trajectories, where the probabilistic nature of quantum mechanics dominates and any particular boundary problem solutions will have to be bounded by.
To understand the difference between basic uncertainty and experimental error, take this example: the line drawn by a compass and a lead pencil, which with mathematics we idealize as a perfect circular trajectory. This is not true "experimentally" as a microscope would show all the little bits of lead dispersed by the pencil . This is due to the experimental construct of the pencil and has an experimental error sigma(x). Since we are talking of a mathematical function we are free to imagine the in reality the circle is perfect on as small errors as one can make. This holds true for the mathematical functions of classical mechanics.
What happens at the quantum mechanical level, when $\hbar$ becomes commensurable with the variables measured, the uncertainty is inherent to the way that nature behaves,
The bubble chamber picture of an electron is a good example. It is curving in a helix in a magnetic field, and the little dots that make up the track are small interactions with the hydrogen atoms, tiny dots of kicked off electrons ionizing even more atoms and making up the dots. The continuous energy loss reduces the radius of the theoretical spiral to the effect seen. This is the macroscopic picture and we call it the trajectory of the electron. We have found out that one cannot go to a microscopic detection that will give a specific trajectory within the HUP volume, the electron is within a fuzzy circle, which is the probability of finding the electron within the circle, the non-zero-probability-of-interaction circle. Thus there are no trajectories in the quantum mechanical frame, just loci of probability for an interaction of the particle under examination.