Is "excess thrust" possible in a steady state vector diagram?
First, we must be clear on what exactly is &quot;excess thrust&quot;. I will list two possible definitions, although many more may be possible.

Excess thrust is the component of the resultant force in the direction of the flight path. Using this definition, excess thrust and steady state flight are directly at odds (because any net force results in an acceleration). So in that case, no, excess thrust is not possible in steady state flight.
Excess thrust is thrust minus drag. It splits the big lumpy thing called 'thrust' in two numbers: one that is used to counter drag (which is always there), and another that can be used for useful things.

Since we already answered the first option, I will now produce the free-body diagrams for the second option.

On the left, we have level flight. We recognize the four fundamental forces of flight: lift (blue), weight (green), drag (red), thrust (orange). It is immediately clear that all forces cancel each other, and we will happily coast along.
In the middle, we have a steady state climb. All vectors, except the weight vector, are 'tilted' back and change magnitude. Lift has decreased slightly, and as a consequence, drag has also decreased minutely (although this doesn't really matter). The biggest change can be seen in thrust, which has almost doubled in magnitude!
On the right image, I have proven that despite all these changes, the net force is still zero. I have replaced the lift- and drag vector in purple by subtracting drag from thrust. But wait a second! That purple vector is exactly what I called 'excess thrust': thrust minus drag! So this purple vector gives us a nice indication of how much 'extra' thrust we are using to make sure the resultant force on the airplane is still zero.
The sometimes heard term, &quot;max excess thrust available&quot; is then an indication of how far we can 'tilt back' the blue lift vector. On the diagram on the right, it should be obvious that this 'stretches' the purple 'excess thrust' vector. Steepest climb is thus achieved when the excess thrust equals the max excess thrust available.
Glossary

Resultant force: also called net force. The sum of all forces that act on the aircraft.
Steady state flight: all acceleration (in Earth's reference frame) is zero. I.e., level flight, steady climb, steady descent. Due to $\vec{F}=m\vec{a}$, zero acceleration means zero resultant force.

Addendum I: From the comments, &quot;A Piper Cub cannot take part of its thrust and apply it for lift.&quot; We see in the image above that sadly, that is exactly what a Piper Cub must be doing to climb at all. The excess thrust $T_e$ requirement $T_e = W \sin \beta$ with $W$ lift and $\beta$ the climb angle. If very little excess thrust is available, we had better keep $\beta$ small - exactly what you would expect for a little cub!
Addendum II: below, the same image but with excess thrust according to definition 1. The vectors are no longer a 'closed triangle', and the net force is not zero. The aircraft is now accelerating.

Images created using IPE.