Call $\alpha\in
On$ *p.r.closed*, if every *p.r. set function* $f$ is total on $L_{\alpha}$. 

(Note (1) If $\alpha^\ast$ is the least p.r. closed  ordinal $> \alpha$ then it is the next image of a point in the $\omega$'th Veblen function, so in any case is much smaller than the next admissible. Hence
(2) the p.r.closed ordinals form a c.u.b. class, and in fact are c.u.b. below any admissible ordinal $>\omega$. (3) By Jensen-Karp [1] the p.r. functions on ordinals to ordinals are all the restrictions of the p.r. set functions to $On$. Hence it is easier to reason with the latter class.)

*Definition*  Let $\delta$ be *p.r. reflecting* if for all p.r. functions $F:On\rightarrow On$, if $F(\delta)\neq 0$ then $\exists \alpha < \delta (F(\alpha)\neq 0 )$.

Then

*Claim* $\delta$ *not p.r. reflecting iff $\delta$ is p.r. recognizable.*

Now let $\beta$ be the least p.r. closed ordinal with a $\beta_0<\beta$ satisfying $L_{\beta_0}\prec_{\Sigma_1}L_\beta$.  (So larger than the least $\Pi^1_0$-reflecting ordinal, but less than  the least $\Pi^1_1$-reflecting ordinal.) By p.r. closure, and $\Sigma_1$-elementarity, the totality of the p.r. functions on $L_\beta$ ensures:

*Lemma The least p.r. reflecting ordinal is less than or equal to $\beta_0$.*

This gives an upper bound to the first p.r. reflecting, and so non-p.r. recognizable, ordinal, and will answer the first part of the Extra Quest. negatively, as we shall see below there are p.r. recognisable ordinals $>\beta$.

As an imprecise  lower bound one can take the least $\Pi^1_0$-reflecting ordinal mentioned above. (If $\gamma$ is this ordinal, and so we have $L_\gamma \prec_{\Sigma_1}L_{\gamma +1}$, it is easy to see all ordinals $\tau < \gamma$ are p.r. recog. as for any such $\tau$ there is some sentence $B$ so that $L_\tau$ is the least level of the $L$-hierarchy where $B$ is true. But $\gamma+1$ is itself p.r. recog. as the first $\tau$ where $L_\tau$ has a proper $\Sigma_1$-substructure, etc., etc. (This answers the second part of the original Question) Similar statements hold for larger stretches of the ordinals $[0,\gamma']$ for $\gamma + 1 < \gamma'$.) 

 
To find the least upper bound of the p.r. recog. ordinals, one can reason as follows:

If the $\Sigma_1$ sentence $A$ is first true at some stage $L_{\delta}$, then $\delta$ is p.r. recog. Consequently 
the p.r. recog. ordinals will be cofinal in $\sigma_1$ the least $\Sigma_1$-stable ordinal, *i.e.* the least ordinal with $L_{\sigma}\prec_{\Sigma_1}V $ (as new $\Sigma_1$-sentences become true cofinally in $\sigma_1$). Thus for example if $\delta$ is least so that $L_\delta$ is a $ZF^-$ model, it will be p.r. recog. Note then that no ordinal $\geq \sigma_1$ can be p.r.recongizable, as otherwise this would be a new $\Sigma_1$ fact true at a stage beyond $\sigma_1$.  Thus:

*Lemma The supremum of the  p.r. recognisable ordinals is $\sigma_1$.*

If $\tau_0$ is the least p.r. reflecting ordinal, and $\tau >\tau_0$ the least p.r. closed ordinal above that, then the answer  to the second question of the Extra Question, is to take $\tau +1$. The $\Sigma_1$ statement that "there exists such a pair $\tau_0<\tau$ " can be used to recognise this ordinal.



Comment: if the definition of recognizable is adjusted (call it recognizable*) so that $\alpha$ is recognizable* if for some p.r. function $F$, $F$ is everywhere $0$ except that for some (unique) $\tau$ $F(\tau)=\alpha$, (and thus $\alpha$ is the sole non-zero value that $F$ takes) then the recognizable* ordinals are precisely those $< \sigma_1$.


[1] R.B. Jensen and C. Karp *Primitive Recursive set functions*, in Proceedings of Symposia in Pure Mathematics,vol.13 Part 1, *"Axiomatic Set Theory"*, Ed. D. Scott, AMS, 1971, pp 143-167.